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>
template <typename T>
void scan(T &x) {
  x = 0;
  bool _ = 0;
  T c = getchar();
  _ = c == 45;
  c = _ ? getchar() : c;
  while (c < 48 || c > 57) c = getchar();
  for (; c < 48 || c > 57; c = getchar())
    ;
  for (; c > 47 && c < 58; c = getchar()) x = (x << 3) + (x << 1) + (c & 15);
  x = _ ? -x : x;
}
template <typename T>
void printn(T n) {
  bool _ = 0;
  _ = n < 0;
  n = _ ? -n : n;
  char snum[65];
  int i = 0;
  do {
    snum[i++] = n % 10 + 48;
    n /= 10;
  } while (n);
  --i;
  if (_) putchar(45);
  while (i >= 0) putchar(snum[i--]);
}
template <typename First, typename... Ints>
void scan(First &arg, Ints &...rest) {
  scan(arg);
  scan(rest...);
}
template <typename T>
void print(T n) {
  printn(n);
  putchar(10);
}
template <typename First, typename... Ints>
void print(First arg, Ints... rest) {
  printn(arg);
  putchar(32);
  print(rest...);
}
using namespace std;
const int MM = 1e5 + 2;
int t;
char s[MM];
int main() {
  scanf("%d", &t);
  while (t--) {
    int n, x, cnt, mx, mn;
    scanf("%d %d %s", &n, &x, s);
    cnt = mx = mn = 0;
    for (int i = 0; i < n; i++) {
      if (s[i] == '0')
        cnt++;
      else
        cnt--;
      mx = max(mx, cnt);
      mn = min(mn, cnt);
    }
    if (cnt == 0) {
      if (x <= mx and x >= mn)
        puts("-1");
      else
        puts("0");
      continue;
    }
    int ans = 0, nv = 0;
    for (int i = 0; i < n; i++) {
      int m = abs(cnt), eq = ((((nv % m) + m) % m) == ((x % m + m) % m));
      if (nv == x)
        ans++;
      else if (eq and ((x >= nv and cnt > 0) or (x < nv and cnt < 0)))
        ans++;
      if (s[i] == '0')
        nv++;
      else
        nv--;
    }
    print(ans);
  }
  return 0;
}