Penalty（逻辑分析）

Consider a simplified penalty phase at the end of a football match.

A penalty phase consists of at most 10 kicks, the first team takes the first kick, the second team takes the second kick, then the first team takes the third kick, and so on. The team that scores more goals wins; if both teams score the same number of goals, the game results in a tie (note that it goes against the usual football rules). The penalty phase is stopped if one team has scored more goals than the other team could reach with all of its remaining kicks. For example, if after the 7-th kick the first team has scored 1goal, and the second team has scored 3 goals, the penalty phase ends — the first team cannot reach 3goals.

You know which player will be taking each kick, so you have your predictions for each of the 10 kicks. These predictions are represented by a string ss consisting of 10 characters. Each character can either be 1, 0, or ?. This string represents your predictions in the following way:

• if sisi is 1, then the ii-th kick will definitely score a goal;
• if sisi is 0, then the ii-th kick definitely won't score a goal;
• if sisi is ?, then the ii-th kick could go either way.

Based on your predictions, you have to calculate the minimum possible number of kicks there can be in the penalty phase (that means, the earliest moment when the penalty phase is stopped, considering all possible ways it could go). Note that the referee doesn't take into account any predictions when deciding to stop the penalty phase — you may know that some kick will/won't be scored, but the referee doesn't.

Input

The first line contains one integer tt (1≤t≤1000) — the number of test cases.

Each test case is represented by one line containing the string ss, consisting of exactly 10 characters. Each character is either 1, 0, or ?.

Output

For each test case, print one integer — the minimum possible number of kicks in the penalty phase.

Example

Input

1. 4
2. 1?0???1001
3. 1111111111
4. ??????????
5. 0100000000

Output

1. 7
2. 10
3. 6
4. 9

Note

Consider the example test:

In the first test case, consider the situation when the 1-st, 5-th and 7-th kicks score goals, and kicks 2, 3, 4 and 6 are unsuccessful. Then the current number of goals for the first team is 3, for the second team is 0, and the referee sees that the second team can score at most 2 goals in the remaining kicks. So the penalty phase can be stopped after the 7-th kick.

In the second test case, the penalty phase won't be stopped until all 10 kicks are finished.

In the third test case, if the first team doesn't score any of its three first kicks and the second team scores all of its three first kicks, then after the 6-th kick, the first team has scored 0 goals and the second team has scored 3 goals, and the referee sees that the first team can score at most 2 goals in the remaining kicks. So, the penalty phase can be stopped after the 6-th kick.

In the fourth test case, even though you can predict the whole penalty phase, the referee understands that the phase should be ended only after the 9-th kick.

题目分析，就是我们可以明确知道是否进球通过0，1，这时候我们可以把？分成a队必进，b队不进，然后计算这种情况，第二种就是b队必进，a队必不进，最后把这2种情况比较大小取小的。

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上ac代码。

有事你就q我；QQ2917366383

学习算法

#include<iostream>
#include<cmath>
using namespace std;
  string s;
inline int zhuq1a()//支持a队
{
  int a1=5,b1=5,a1s=0,b1s=0;//a1表示a队剩下的进球机会，b1同理，a1s则是a队已经得到的分数，b1s同理
  for(int i=0;i<=9;i++)
  {
    if(i%2==0)
    {a1--;//别忘了处理剩下的进球机会
    if(s[i]=='?')
    a1s++;  //默认进球    
    else 
    a1s+=(int)s[i]-'0';//字符转换为数字
      }
    else
    {     
      b1--;
      if(s[i]=='?')
      {//什么都不做，即默认不进球
       } 
       else
       b1s+=(int)s[i]-'0';
     } 
     if(b1s+b1<a1s)return i+1;//这里i是字符串下标，转换成点球的轮数需要+1    
  }
  return 10;
}
inline int zhuq1b()//支持b队，同上
{
  int a1=5,b1=5,a1s=0,b1s=0;
  for(int i=0;i<=9;i++)
  {
    if(i%2==0)
    {a1--;
    if(s[i]=='?')
    {}    
    else 
    a1s+=(int)s[i]-'0';
      }
    else
    {     
      b1--;
      if(s[i]=='?')
      {
        b1s++;
       } 
       else
       b1s+=(int)s[i]-'0';
     } 
     if(a1s+a1<b1s)return i+1;    
  }
  return 10;
}
int main()
{
  int t;
  cin>>t;
  while(t--)//CF多组数据
  {
        cin>>s;
cout<<min(zhuq1a(),zhuq1b())<<endl;
  }
}