# B. Infinite Prefixes

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

Calculate the number of prefixes of t with balance equal to x. The balance of some string q is equal to cnt0,qcnt1,q, where cnt0,q is the number of occurrences of 0 in q, and cnt1,q is the number of occurrences of 1 in q. 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 T (1 ≤ T ≤ 100)— the number of test cases.

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

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

It’s guaranteed that the total sum of n doesn’t exceed 105.

## Output

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

## Example

### input

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


### output

3
0
1
-1


## Note

In the first test case, there are 3 good prefixes of t: with length 28, 30 and 32.

# Problem solving:

num 不为 0

s = 010010 , x = 10

AC代码：

#include<bits/stdc++.h>
#define fi        first
#define se        second
#define PI        acos(-1)
#define LC(a)     ((a<<1))
#define RC(a)     ((a<<1)+1)
#define MID(a,b)  ((a+b)>>1)
#define mem(a, b) memset(a, b, sizeof(a))
#define IOS()     std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
using namespace std;
typedef long long ll;
typedef unsigned long long ULL;
typedef pair<int,int> PII;
typedef pair<ll,ll>   PLL;
typedef pair<ll,int>  PLI;
const int INF = 0X3F3F3F3F;
const int MIN = -(1<<30);
const ll N = 1e6+7;
ll a[N];
int main()
{
IOS();
int t;
cin>>t;
while(t--)
{
ll n,x;
string s;
cin>>n>>x>>s;
if(x<0)
{
x=-x;
for(int i=0; i<n; i++)
s[i]=s[i]=='0'?'1':'0';
}
//        cout<<s<<endl;
ll num=0;
for(int i=0; i<s.size(); i++)
{
if(s[i]=='0') num++;
else num--;
a[i]=num;
}
ll ans=0;
int flag = 0;

if(num==0)
{
for(int i=0; i<n; i++)
{
if(a[i]==x) flag=1;
}
}
else
{
for(int i=0; i<n; i++)
{
if((x-a[i])%num==0 && (x-a[i])/num>=0) ans++;
}
}
if(x==0) ans++;
if(flag) cout<<-1<<endl;
else cout<<ans<<endl;

}
return 0;
}