链接：https://codeforces.com/contest/1295/problem/B

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.

题解：赛时太菜，没想过负数（这好像是第二次掉入负数的坑了），赛后面向样例编程QAQ

代码：

#include <bits/stdc++.h>
using namespace std;
long long l,r,n,t,s0,s1,s,ans,k,max1=0;
long long a,b;
char x[10000001];
long long p[10000001];
int main()
{
	cin>>t;
	while(t--)
	{
		cin>>a>>b;
		cin>>x;
		if(b<0)
		{
			b=-b;
			for(int i=0;i<a;i++)
			{
				if(x[i]=='0')
				x[i]='1';
				else
				x[i]='0';
			}
		}
		s=0;
		ans=0;
		int flag=0;
		for(int i=0;i<a;i++)
		{
			if(x[i]=='0')
			s++;
			else
			s--;
			p[i]=s;
			if(s==b)
			{
				flag=1;
			}
		}
		if(b==0)
		ans++;
		if(flag==0&&s==0)
		{
			cout<<0<<endl;
			continue;
		}
		if(flag==1&&s==0)
		{
			cout<<-1<<endl;
			continue;
		}
		if(s!=0)
		{
			for(int i=0;i<a;i++)
			{
				if(b==p[i])
				ans++;
				else
				{
					if(b>p[i]&&s>0)
					{
						if((b-p[i])%s==0)
						ans++;
					}
					else if(b<p[i]&&s<0)
					{
						if((b-p[i])%s==0)
						ans++;
					}
				}
			}
		}
		cout<<ans<<endl;
	}
}