B. Infinite Prefixes

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.

 

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <queue>
#include <map>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <numeric>
#include <cmath>
#include <iomanip>
#include <deque>
#include <bitset>
//#include <unordered_set>
//#include <unordered_map>
//#include <bits/stdc++.h>
//#include <xfunctional>
#define ll              long long
#define PII             pair<int, int>
#define rep(i,a,b)      for(int  i=a;i<=b;i++)
#define dec(i,a,b)      for(int  i=a;i>=b;i--)
#define pb              push_back
#define mk              make_pair
using namespace std;
int dir1[6][2] = { { 0,1 } ,{ 0,-1 },{ 1,0 },{ -1,0 },{ 1,1 },{ -1,1 } };
int dir2[6][2] = { { 0,1 } ,{ 0,-1 },{ 1,0 },{ -1,0 },{ 1,-1 },{ -1,-1 } };
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int inf = 0x3f3f3f3f;
const double pi = 3.14159265358979;
const int mod = 100007;
const int N = 1005;
//if(x<0 || x>=r || y<0 || y>=c)

inline ll read()
{
    ll x = 0; bool f = true; char c = getchar();
    while (c < '0' || c > '9') { if (c == '-') f = false; c = getchar(); }
    while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
    return f ? x : -x;
}
ll gcd(ll m, ll n)
{
    return n == 0 ? m : gcd(n, m % n);
}

int main()
{
    int T;
    cin >> T;
    while (T--)
    {
        ll n, x;
        cin >> n >> x;
        string s;
        cin >> s;
        vector<ll> a(n+1,0);
        ll res=0;
        if (x == 0)
            res++;
        for (int i = 0; i < s.size(); i++)
        {
            if (s[i] == '0')
            {
                a[i + 1] = a[i] + 1;
            }
            else
            {
                a[i + 1] = a[i] - 1;
            }
            if (a[i + 1] == x)
                res++;
        }
        if (a[n] == 0 && res > 0)
        {
            cout << -1 << endl;
            continue;
        }
        else
        {
            if (a[n] == 0)
            {
                cout << res << endl;
                continue;
            }
            rep(i, 1, n)
            {
                if ((x-a[i])% a[n]==0 && (x - a[i]) / a[n] > 0)
                    res++;
            }
            cout << res << endl;
        }
    }
    return 0;
}