Problem: 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 cnt_{0, q} - cnt_{1, q}, where cnt_{0, q} is the number of occurrences of 0 in q, and cnt_{1, 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 ≤ 10^5, -10^9 ≤ x ≤ 10^9) — the length of string s and the desired balance, respectively.

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

It's guaranteed that the total sum of n doesn't exceed 10^5.

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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
long long NSD(long long a, long long b) {
  while (a != 0 && b != 0) {
    if (a > b) {
      a %= b;
    } else {
      b %= a;
    }
  }
  return (a + b);
}
long long M = INFINITY;
long long multipleByMod(long long a, long long b) {
  return (a * b) % M;
  if (!b) {
    return 0;
  }
  a %= M;
  b %= M;
  long long c = multipleByMod(a, b / 2) * 2;
  c %= M;
  if (b & 1) {
    return (c + a) % M;
  } else {
    return c;
  }
}
long long step(long long a, long long b) {
  if (!b) return 1;
  long long c = step(a, b / 2) % M;
  c = multipleByMod(c, c);
  if (b & 1)
    return multipleByMod(c, a);
  else
    return c;
}
using namespace std;
mt19937 rnd(time(nullptr));
void sol() {
  long long n, x;
  cin >> n >> x;
  string s;
  cin >> s;
  if (x < 0) {
    x = -x;
    for (auto &B : s) {
      B = B + 1 - 2 * (B == '1');
    }
  }
  int cnt[] = {0, 0};
  for (auto &B : s) {
    cnt[B - '0']++;
  }
  if (cnt[0] - cnt[1] == 0 && x == 0) {
    cout << "-1\n";
    return;
  }
  long long ans = (x == 0), ma = 0, res = 0;
  for (auto &B : s) {
    if (B == '0') {
      res++;
    } else {
      res--;
    }
    if (cnt[0] == cnt[1]) {
      if (x == res) {
        cout << "-1\n";
        return;
      }
    } else {
      if ((x - res) % (cnt[0] - cnt[1]) == 0 &&
          (x - res) * (cnt[0] - cnt[1]) >= 0) {
        ans++;
      }
    };
    ;
    ma = max((long long)(ma), (long long)(res));
  }
  if (cnt[0] - cnt[1] <= 0 && ma < x) {
    cout << "0\n";
    return;
  }
  cout << ans << "\n";
}
int main() {
  ;
  ;
  cin.tie(nullptr);
  ios_base::sync_with_stdio(false);
  cout.tie(nullptr);
  ;
  int t;
  cin >> t;
  while (t--) {
    sol();
  }
}