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 collections import defaultdict as d

t=int(input())
for _ in range(t):
    n,x=[int(aa) for aa in input().split()]
    s=input()
    
    balanceCounts=d(lambda:0) #balance for 1 cycle. {intermediate bal:counts}
    balanceCounts[0]+=1 #empty prefix
    bal=0
    for i in range(n):
        if s[i]=='1':
            bal-=1
        else:
            bal+=1
        balanceCounts[bal]+=1
    #bal is now the change in balance per cycle
    balanceCounts[bal]-=1 #let the full cycle be taken care of by the next cycle to avoid double counting
#    print(balanceCounts,bal)
    if bal==0:
        if balanceCounts[x]>0:
            ans=-1 #infinite prefixes
        else:
            ans=0 #impossible
    else:
        minBal=min(balanceCounts.keys())
        maxBal=max(balanceCounts.keys())
        
        #use k*bal + minBal to maxBal = x
#        if bal>0:
#            kUpper=max(0,1+(x-minBal)//bal)
#            kLower=max(0,(x-maxBal)//bal)
#        else:
#            kUpper=max(0,1+(x-maxBal)//bal)
#            kLower=max(0,(x-maxBal)//bal)
        aa=max(0,(x-minBal)//bal)
        bb=max(0,(x-maxBal)//bal)
        kLower=min(aa,bb)
        kUpper=1+max(aa,bb)

#        print('balanceCounts:{} bal:{} kUpper:{} klower:{}'.format(balanceCounts,bal,kUpper,kLower))
        ans=0
        for k in range(kLower,kUpper+1):
            intermediateBalance=x-k*bal
            ans+=balanceCounts[intermediateBalance]
#    print('here')
    print(ans)