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>
using namespace std;
int find_mx1(string s) {
  int cnt = 0;
  vector<int> temp;
  for (int i = 0; i < s.size(); i++) {
    if (s[i] == '0')
      cnt++;
    else
      cnt--;
    temp.push_back(cnt);
  }
  return *max_element(temp.begin(), temp.end());
}
int find_mx2(string s) {
  int cnt = 0;
  vector<int> temp;
  for (int x = 1; x <= 2; x++)
    for (int i = 0; i < s.size(); i++) {
      if (s[i] == '0')
        cnt++;
      else
        cnt--;
      temp.push_back(cnt);
    }
  return *max_element(temp.begin(), temp.end());
}
int find_mn1(string s) {
  int cnt = 0;
  vector<int> temp;
  for (int i = 0; i < s.size(); i++) {
    if (s[i] == '0')
      cnt++;
    else
      cnt--;
    temp.push_back(cnt);
  }
  return *min_element(temp.begin(), temp.end());
}
int find_mn2(string s) {
  int cnt = 0;
  vector<int> temp;
  for (int x = 1; x <= 2; x++)
    for (int i = 0; i < s.size(); i++) {
      if (s[i] == '0')
        cnt++;
      else
        cnt--;
      temp.push_back(cnt);
    }
  return *min_element(temp.begin(), temp.end());
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n, x;
    cin >> n >> x;
    string s;
    cin >> s;
    int mx1 = find_mx1(s);
    int mn1 = find_mn1(s);
    int mx2 = find_mx2(s);
    int mn2 = find_mn2(s);
    if (mx2 > mx1) {
      if (x < mn1) {
        if (x == 0)
          cout << "1" << endl;
        else
          cout << "0" << endl;
      } else {
        int gap = mx2 - mx1;
        vector<int> temp;
        int cnt = 0;
        for (int i = 0; i < s.size(); i++) {
          if (s[i] == '0')
            cnt++;
          else
            cnt--;
          temp.push_back(cnt);
        }
        int ans = 0;
        for (int i = 0; i < temp.size(); i++) {
          if ((x - temp[i]) % gap == 0 && (x - temp[i]) / gap >= 0) ans++;
        }
        if (x == 0) ans++;
        cout << ans << endl;
      }
    } else {
      if (mn1 == mn2) {
        if (x >= mn1 && x <= mx1)
          cout << "-1" << endl;
        else {
          if (x == 0)
            cout << "1" << endl;
          else
            cout << "0" << endl;
        }
      } else if (mn2 < mn1) {
        if (x > mx1) {
          if (x == 0)
            cout << "1" << endl;
          else
            cout << "0" << endl;
        } else {
          int gap = mn2 - mn1;
          int cnt = 0;
          vector<int> temp;
          for (int i = 0; i < s.size(); i++) {
            if (s[i] == '0')
              cnt++;
            else
              cnt--;
            temp.push_back(cnt);
          }
          int ans = 0;
          for (int i = 0; i < temp.size(); i++) {
            if ((x - temp[i]) % gap == 0 && (x - temp[i]) / gap >= 0) ans++;
          }
          if (x == 0) ans++;
          cout << ans << endl;
        }
      }
    }
  }
}