CF1288A Deadline（枚举）

Adilbek was assigned to a special project. For Adilbek it means that he has nn days to run a special program and provide its results. But there is a problem: the program needs to run for  d days to calculate the results.

Fortunately, Adilbek can optimize the program. If he spends x  ( x is a non-negative integer) days optimizing the program, he will make the program run in ⌈dx+1⌉  days (⌈a⌉ is the ceiling function: ⌈2.4⌉=3，⌈2⌉=2 ). The program cannot be run and optimized simultaneously, so the total number of days he will spend is equal to x+⌈dx+1⌉

Will Adilbek be able to provide the generated results in no more than nn days?

Input

The first line contains a single integer T (1≤T≤50 ) — the number of test cases.

The next T  lines contain test cases – one per line. Each line contains two integers n and d  (1≤n≤10^9 1≤d≤10^9 ) — the number of days before the deadline and the number of days the program runs.

Output

Print  T answers — one per test case. For each test case print YES (case insensitive) if Adilbek can fit in nn days or NO (case insensitive) otherwise.

Example

Input

1. 3
2. 1 1
3. 4 5
4. 5 11

Output

1. YES
2. YES
3. NO

Note

In the first test case, Adilbek decides not to optimize the program at all, since d≤n

In the second test case, Adilbek can spend 1  day optimizing the program and it will run ⌈5/2⌉=3 days. In total, he will spend  4 days and will fit in the limit.

In the third test case, it's impossible to fit in the limit. For example, if Adilbek will optimize the program 2  days, it'll still work ⌈11/2+1⌉=4  days.

题目分析，我们这个题首先我们可以看到是天花板函数当相除都时候，所以等下我们枚举的时候要处理一下，然后遍历找答案就行了，具体实现看代码。

#include<iostream>
using namespace std;
int main()
{
  int t;
  cin>>t;
  while(t--)
  {
    int n,m;int a=0;
  cin>>n>>m;
  for(int i=0;i<=n;i++)
  {
    if((m-1)/(i+1)+1+i<=n)//解释因为·这里要的是天花板函数，所以如果有小数要加1； 
    {
      a=1;
      cout<<"YES"<<endl;
      break;
    }
   }
   if(!a)
    cout<<"NO"<<endl;
  } 
}