Educational Codeforces Round 82 A. Erasing Zeroes

You are given a string ss. Each character is either 0 or 1.

You want all 1's in the string to form a contiguous subsegment. For example, if the string is 0, 1, 00111 or 01111100, then all 1's form a contiguous subsegment, and if the string is 0101, 100001 or 11111111111101, then this condition is not met.

You may erase some (possibly none) 0's from the string. What is the minimum number of 0's that you have to erase?

Input

The first line contains one integer tt (1≤t≤1001≤t≤100) — the number of test cases.

Then tt lines follow, each representing a test case. Each line contains one string ss (1≤|s|≤1001≤|s|≤100); each character of ss is either 0 or 1.

Output

Print tt integers, where the ii-th integer is the answer to the ii-th testcase (the minimum number of 0's that you have to erase from ss).

Example

Input

 

3
010011
0
1111000

Output

 

2
0
0
大意就是问最少删除多少个给定序列里的0能让所有的1都毗连。不妨统计所有1的位置并遍历，发现如果两个1的位置下标的差大于1，则更新答案（不要忘记特判）。

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        char s[105];
        scanf("%s",s);
        int i;
        vector<int>v;
        int ans=0;
        for(i=0;i<strlen(s);i++)
        {
            if(s[i]=='1')v.push_back(i);
        }
        if(v.size()==1||v.size()==0||strlen(s)==1)
        {
            cout<<0<<endl;
            continue;
        }
        for(i=0;i<v.size()-1;i++)
        {
            if(v[i+1]-v[i]!=1)ans+=(v[i+1]-v[i]-1);
        }
        cout<<ans<<endl;
    }
    return 0;
}