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;
const int maxn = 2e5 + 100;
const long long mod = 998244353;
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const double eps = 1e-7;
long long GCD(long long a, long long b) {
  while (b ^= a ^= b ^= a %= b)
    ;
  return a;
}
long long LCM(long long a, long long b) { return a * b / GCD(a, b); }
long long prime[maxn] = {0}, num_prime = 0;
int isNotPrime[maxn] = {1, 1};
inline long long read() {
  long long X = 0, w = 0;
  char ch = 0;
  while (!isdigit(ch)) {
    w |= ch == '-';
    ch = getchar();
  }
  while (isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
  return w ? -X : X;
}
inline void write(int x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
void isPrime() {
  for (int i = 2; i < maxn; ++i) {
    if (!isNotPrime[i]) prime[num_prime++] = i;
    for (int j = 0; j < num_prime && i * prime[j] < maxn; ++j) {
      isNotPrime[i * prime[j]] = 1;
      if (!i % prime[j]) break;
    }
  }
  printf("%lld", prime[num_prime - 1]);
}
long long binaryPow(long long a, long long b, long long c) {
  long long ans = 1;
  while (b > 0) {
    if (b & 1) {
      ans = ans * a % c;
    }
    a = a * a % c;
    b >>= 1;
  }
  return ans % c;
}
long long T, n, x;
char str[maxn];
int arr[maxn];
int main() {
  scanf("%lld", &T);
  while (T--) {
    memset(str, 0, sizeof(str));
    scanf("%lld %lld", &n, &x);
    scanf("%s", &str);
    int cnt0 = 0, cnt1 = 0;
    int ans = 0;
    int val = 0;
    for (int i = 0; i < n; ++i) {
      if (str[i] == '0')
        cnt0++;
      else
        cnt1++;
      arr[i] = cnt0 - cnt1;
      if (arr[i] == x) val++;
    }
    if (arr[n - 1] == 0 && val > 0) {
      printf("-1\n");
      continue;
    }
    if (arr[n - 1] == 0)
      ans = val;
    else if (arr[n - 1] > 0) {
      for (int i = 0; i < n; ++i) {
        if ((x - arr[i] >= 0) && (x - arr[i]) % arr[n - 1] == 0) ans++;
      }
    } else if (arr[n - 1] < 0) {
      for (int i = 0; i < n; ++i) {
        if ((x - arr[i] <= 0) && (x - arr[i]) % arr[n - 1] == 0) ans++;
      }
    }
    if (x == 0) ans++;
    printf("%d\n", ans);
  }
  return 0;
}