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:
from math import *
t=int(input())
while(t):
    t-=1
    n,x=map(int,input().split())
    s=input()
    ref=[]
    z,o=0,0
    for i in s:
        if(i=='0'):
            z+=1
            ref.append(z-o)
        else:
            o+=1
            ref.append(z-o)
    ref1=[]
    s1=0
    if(x==0):
        s1+=1
    flag=0
    for i in s:
        if(i=='0'):
            z+=1
            ref1.append(z-o)
        else:
            o+=1
            ref1.append(z-o)
    di=ref1[0]-ref[0]
    """if(di<0):
        print(0)
        continue"""
    #print(di,ref)
    for i in range(n):
        f=x-ref[i]
        if(di!=0):
            f=f/di
        if(ceil(f)==floor(f) and di!=0):
            f=f+1
            if(f>0):
                s1+=1
        if(di==0 and x==ref[i]):
            flag=1
            break
    if(flag==1):
        print(-1)
    else:
        print(s1)