Codeforces Round 889 (Div. 2)

本次比赛链接：https://codeforces.com/contest/1849

A. Morning Sandwich

Monocarp always starts his morning with a good ol' sandwich. Sandwiches Monocarp makes always consist of bread, cheese and/or ham.

A sandwich always follows the formula:

• a piece of bread
• a slice of cheese or ham
• a piece of bread
• ……
• a slice of cheese or ham
• a piece of bread

So it always has bread on top and at the bottom, and it alternates between bread and filling, where filling is a slice of either cheese or ham. Each piece of bread and each slice of cheese or ham is called a layer.

Today Monocarp woke up and discovered that he has bb pieces of bread, cc slices of cheese and hh slices of ham. What is the maximum number of layers his morning sandwich can have?

Input

The first line contains a single integer tt (1≤t≤10001≤t≤1000) — the number of testcases.

Each testcase consists of three integers b,cb,c and hh (2≤b≤1002≤b≤100; 1≤c,h≤1001≤c,h≤100) — the number of pieces of bread, slices of cheese and slices of ham, respectively.

Output

For each testcase, print a single integer — the maximum number of layers Monocarp's morning sandwich can have.

Example

Example

input

1. 3
2. 2 1 1
3. 10 1 2
4. 3 7 8

output

1. 3
2. 7
3. 5

Note

In the first testcase, Monocarp can arrange a sandwich with three layers: either a piece of bread, a slice of cheese and another piece of bread, or a piece of bread, a slice of ham and another piece of bread.

In the second testcase, Monocarp has a lot of bread, but not too much filling. He can arrange a sandwich with four pieces of bread, one slice of cheese and two slices of ham.

In the third testcase, it's the opposite — Monocarp has a lot of filling, but not too much bread. He can arrange a sandwich with three pieces of bread and two slices of cheese, for example.

思路：除去第一片面包，剩下的面包片数量等于料的数量，直接找到最小值乘二加上第一片面包

AC代码：

#include<bits/stdc++.h>
using namespace std;
#define int long long
#define IOS ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
void solve()
{
  int m,a,b,maxx=0,aa=0;
  cin>>m>>a>>b;
  aa=m-1;
  maxx=min(aa,a+b);
  cout<<maxx*2+1<<endl;
    return ;
}
signed main()
{
  int t=1;
  cin>>t;
  while(t--) solve();
  return 0;
}

B. Monsters

Monocarp is playing yet another computer game. And yet again, his character is killing some monsters. There are nn monsters, numbered from 11 to nn, and the ii-th of them has aiai health points initially.

Monocarp's character has an ability that deals kk damage to the monster with the highest current health. If there are several of them, the one with the smaller index is chosen. If a monster's health becomes less than or equal to 00 after Monocarp uses his ability, then it dies.

Monocarp uses his ability until all monsters die. Your task is to determine the order in which monsters will die.

Input

The first line contains a single integer tt (1≤t≤1041≤t≤104) — the number of test cases.

The first line of each test case contains two integers nn and kk (1≤n≤3⋅1051≤n≤3⋅105; 1≤k≤1091≤k≤109) — the number of monsters and the damage which Monocarp's ability deals.

The second line contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤109) — the initial health points of monsters.

The sum of nn over all test cases doesn't exceed 3⋅1053⋅105.

Output

For each test case, print nn integers — the indices of monsters in the order they die.

Example

input

1. 3
2. 3 2
3. 1 2 3
4. 2 3
5. 1 1
6. 4 3
7. 2 8 3 5

output

1. 2 1 3
2. 1 2
3. 3 1 2 4

Note

In the first example, the health points change as follows: [1,2,3–]→[1,2–,1]→[1–,0,1]→[−1,0,1–]→[−1,0,−1][1,2,3_]→[1,2_,1]→[1_,0,1]→[−1,0,1_]→[−1,0,−1]. The monster that is going to take damage the next time Monocarp uses his ability is underlined.

In the second example, the health points change as follows: [1–,1]→[−2,1–]→[−2,−2][1_,1]→[−2,1_]→[−2,−2].

In the third example, the health points change as follows: [2,8–,3,5]→[2,5–,3,5]→[2,2,3,5–]→[2,2,3–,2]→[2–,2,0,2]→[−1,2–,0,2]→[−1,−1,0,2–]→[−1,−1,0,−1][2,8_,3,5]→[2,5_,3,5]→[2,2,3,5_]→[2,2,3_,2]→[2_,2,0,2]→[−1,2_,0,2]→[−1,−1,0,2_]→[−1,−1,0,−1].

思路：将每个血量模上伤害值后放进结构体数组，对于能整除伤害值的不做操作，然后按从大到小排一下序，输出对应下标即可

AC代码：

#include<bits/stdc++.h>
using namespace std;
#define int long long
#define IOS ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
struct s{
  int x,y;
}a[310000];
bool cmp(s aa,s bb)
{
  if(aa.x==bb.x)
  return aa.y<bb.y;
  else
  return aa.x>bb.x;
}
void solve()
{
  int n,k;
  scanf("%lld%lld",&n,&k);
  for(int i=1;i<=n;i++)
  {
  scanf("%lld",&a[i].x);
  if(a[i].x%k!=0)
  a[i].x=a[i].x%k;
  else
  a[i].x=k;
  a[i].y=i;
  }
  sort(a+1,a+1+n,cmp);
  for(int i=1;i<=n;i++)
  printf("%lld ",a[i].y);
  printf("\n");
    return ;
}
signed main()
{
  int t;
  scanf("%lld",&t);
  while(t--) solve();
  return 0;
}

C. Binary String Copying

You are given a string ss consisting of nn characters 0 and/or 1.

You make mm copies of this string, let the ii-th copy be the string titi. Then you perform exactly one operation on each of the copies: for the ii-th copy, you sort its substring [li;ri][li;ri] (the substring from the lili-th character to the riri-th character, both endpoints inclusive). Note that each operation affects only one copy, and each copy is affected by only one operation.

Your task is to calculate the number of different strings among t1,t2,…,tmt1,t2,…,tm. Note that the initial string ss should be counted only if at least one of the copies stays the same after the operation.

Input

The first line contains a single integer tt (1≤t≤1041≤t≤104) — the number of test cases.

The first line of each test case contains two integers nn and mm (1≤n,m≤2⋅1051≤n,m≤2⋅105) — the length of ss and the number of copies, respectively.

The second line contains nn characters 0 and/or 1 — the string ss.

Then mm lines follow. The ii-th of them contains two integers lili and riri (1≤li≤ri≤n1≤li≤ri≤n) — the description of the operation applied to the ii-th copy.

The sum of nn over all test cases doesn't exceed 2⋅1052⋅105. The sum of mm over all test cases doesn't exceed 2⋅1052⋅105.

Output

Print one integer — the number of different strings among t1,t2,…,tmt1,t2,…,tm.

Example

input

1. 3
2. 6 5
3. 101100
4. 1 2
5. 1 3
6. 2 4
7. 5 5
8. 1 6
9. 6 4
10. 100111
11. 2 2
12. 1 4
13. 1 3
14. 1 2
15. 1 1
16. 0
17. 1 1

output

1. 3
2. 3
3. 1

Note

Consider the first example. Copies below are given in order of the input operations. Underlined substrings are substrings that are sorted:

101100 →→ 011100;

101100 →→ 011100;

101100 →→ 101100;

101100 →→ 101100;

101100 →→ 000111.

There are three different strings among t1,t2,t3,t4,t5t1,t2,t3,t4,t5: 000111, 011100 and 101100.

Consider the second example:

100111 →→ 100111;

100111 →→ 001111;

100111 →→ 001111;

100111 →→ 010111.

There are three different strings among t1,t2,t3,t4t1,t2,t3,t4: 001111, 010111 and 100111.

AC代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=2e5+10;
ll l[N],r[N];
void work(){
  int n,m;
  string s;
  cin>>n>>m;
  cin>>s;
  s=' '+s;
  l[0]=0;r[n+1]=n+1;
  for(int i=1;i<=n;i++)
  if(s[i]=='1') l[i]=l[i-1];
  else l[i]=i;
  for(int i=n;i>=1;i--)
  if(s[i]=='1') r[i]=i;
  else r[i]=r[i+1];
  set<pair<ll,ll>> se;
  int x,y;
  while(m--){
  cin>>x>>y;
  x=r[x];
  y=l[y];
  if(x>=y) x=y=-1;
  se.insert({x,y});
  }
  printf("%d\n",se.size());
}
int main(){
  int t;
  cin>>t;
  while(t--)
  work();
  return 0;
}