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 o in range(T) :
    n,d=map(int,input().split());s=input()
    num=[0]*(2*n+1);now=0
    

    for i in s :
        if i=='0' : num[now+n]+=1;now+=1
        elif i=='1' : num[now+n]+=1;now-=1
    
    if now==0 :
        if d<=n and d>=-n and num[d+n]>0 : print(-1)
        else : print(0)
    else :
        if (d>n and now<0) or (d<-n and now>0) : print(0)
        elif d>n or d<-n : 
            ans=0
            if now<0 : now=-now
            d%=now;d=(d+now)%now

            for i in range(-n,n) : 
                if (i%now+now)%now==d : ans+=num[i+n]
            print(ans)
        else : 
            ans=0
            while d>=-n and d<=n : 
                ans+=num[d+n];d-=now
            print(ans)