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 = map(int,input().split())
    b = list(input().strip())
    cnt0,cnt1 = [0],[0]
    c0,c1 = 0,0
    for i in b:
        if i=='0':
            c0 += 1
            cnt0.append(cnt0[-1]+1)
            cnt1.append(cnt1[-1])
        else:
            c1 += 1
            cnt0.append(cnt0[-1])
            cnt1.append(cnt1[-1]+1)
    diff = []
    for i in range(n+1):
        diff.append(cnt0[i]-cnt1[i])
    #print(diff)
    if diff[0]==diff[-1]:
        if x in diff:
            print(-1)
        else:
            print(0)
    else:
        k = 0
        inc = diff[-1]
        # if inc*x<0:
        #     print(0)
        inv = []
        for i in diff[:-1]:
            inv.append(x-i)
        #print(inv)
        for i in inv:
            if (i*inc>=0) and (i%inc==0):
                k += 1
        print(k)