Educational Codeforces Round 81 (Rated for Div. 2)B

B. Infinite Prefixes

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given string ss of length nn consisting of 0-s and 1-s. You build an infinite string tt as a concatenation of an infinite number of strings ss, or t=ssss…t=ssss… For example, if s=s= 10010, then t=t= 100101001010010...

Calculate the number of prefixes of tt with balance equal to xx. The balance of some string qq is equal to cnt0,q−cnt1,qcnt0,q−cnt1,q, where cnt0,qcnt0,q is the number of occurrences of 0 in qq, and cnt1,qcnt1,q is the number of occurrences of 1 in qq. The number of such prefixes can be infinite; if it is so, you must say that.

A prefix is a string consisting of several first letters of a given string, without any reorders. An empty prefix is also a valid prefix. For example, the string "abcd" has 5 prefixes: empty string, "a", "ab", "abc" and "abcd".

Input

The first line contains the single integer TT (1≤T≤1001≤T≤100) — the number of test cases.

Next 2T2T lines contain descriptions of test cases — two lines per test case. The first line contains two integers nn and xx (1≤n≤1051≤n≤105, −109≤x≤109−109≤x≤109) — the length of string ss and the desired balance, respectively.

The second line contains the binary string ss (|s|=n|s|=n, si∈{0,1}si∈{0,1}).

It's guaranteed that the total sum of nn doesn't exceed 105105.

Output

Print TT integers — one per test case. For each test case print the number of prefixes or −1−1 if there is an infinite number of such prefixes.

Example

input

Copy

4
6 10
010010
5 3
10101
1 0
0
2 0
01

output

Copy

3
0
1
-1

Note

In the first test case, there are 3 good prefixes of tt: with length 2828, 3030 and 3232.

水题

题意是一个一阶导数为存在周期的函数经过多少次零点

给出一个代表间断函数的导数序列

先计算出周期的sum和

然后判断每个点-x是否能经过零点

注意判断无穷时

代码

//AUTHOR : cwb
#include
using namespace std;
char s[100005];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int ans=0;
        int n,x;
        scanf("%d%d",&n,&x);
        if(x==0) ans++;
        scanf("%s",s+1);
        int sum=0;
        int num=0;
        for(int i=1;s[i];i++)
        {
            if(s[i]=='0') sum+=1;
            else sum-=1;
        }
        for(int i=1;s[i];i++)
        {
            if(s[i]=='0') num++;
            else num--;
            if(num==x) ans++;
            else if(num0&&(x-num)%sum==0) ans++;
            else if(num>x&&sum<0&&(x-num)%sum==0) ans++;
            //printf("%d\n",ans);
        }
        if(sum==0&&ans!=0) printf("-1\n");
        else printf("%d\n",ans);
    }
}