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 PYTHON3):

Solution:
t = int(input())

for _ in range(t):
  n, x = list(map(int, input().split()))
  s = list(map(int, input()))
  bal = [0] * (n + 1)
  cnt = 1 if x == 0 else 0
  for i in range(n):
    if s[i] == 0:
      bal[i+1] = bal[i] + 1
    else:
      bal[i+1] = bal[i] -1
    if bal[i+1] == x:
      cnt += 1
  #print(bal)
  b = bal[-1]
  if b == 0 and cnt > 0:
    print(-1)
  elif b == 0 and cnt == 0:
    print(0)
  elif b != 0:
    for i in range(n):
        rest = x - bal[i+1]
        k = rest // b
        #print(x, bal[i+1], rest, rest % b, k, bal[i+1] + k * b)
        if rest % b == 0 and k > 0:
            cnt += 1
    print(cnt)
  else:
    print(cnt)