CodeForces - 1295B 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

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 tt: with length 2828, 3030 and 3232.

一道水题 模拟

#include
#define ll long long
const int maxn = 1e5 + 5;
using namespace std;
void solve(){
    ll n, x;
    cin >> n >> x;
    char a[maxn];
    ll sum = 0;
    ll f = 0;
    for(ll i = 1; i <= n; i++){
        cin >> a[i];
        if(a[i] == '0')
            sum++;
        else
            sum--;
        if(sum == 0)
            f++;

    }
    if(x == 0){
        if(sum == 0)
            cout << "-1" << endl;
        else{
            if(f == 0)
                cout << 1 << endl;
            else{
                int y = 0;
            int cnt1 = 0;
                for(int i = 1; i <= n; i++){
                    if(a[i] == '0')
                        y++;
                    else
                        y--;
                    if((x - y) % sum == 0 && (x - y) / sum >= 0){
                        cnt1++;
                }

            }
            cout << cnt1 + 1 << endl;
        }
        }
    }
    else{
        ll cnt = 0;
        ll su = 0;
        int fl = 0;
        for(ll i = 1; i <= n; i++){
            if(a[i] == '0')
                su++;
            else
                su--;
            if(sum == 0){
                if(x - su == 0){
                    fl = 1;
                }
            }
            else{
                if((x - su) % sum == 0 && (x - su) / sum >= 0){
                    cnt++;
                }
                /*if((x + su) % sum == 0 && (x + su) / sum >= 0){
                    cnt++;
                }*/
            }
        }
        if(fl)
            cout << -1 << endl;
        else
            cout << cnt << endl;
    }
}
int main(){
    ll t;
    cin >> t;
    while(t--){
        solve();
    }
}