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 i in range(T):
    line=str(input()).split(" ")
    len_string=int(line[0])
    desired_balance=int(line[1])
    string=str(input())


    string_balance=[]
    balance=0
    for el in string:
        if el=="0":
            balance+=1
        else:
            balance-=1
        string_balance.append(balance)

    periodic_addition=string_balance[-1]
    numof_prefixes=0

    if desired_balance==0:
        # i prazdny prefix se pocita, prazdny prefix ma balance=0
        numof_prefixes+=1

    if periodic_addition==0 and desired_balance in string_balance:
        #nekonecne mnozstvi prefixu
        print(-1, end="\n")
        continue

    elif periodic_addition==0:
        # jestli nejaky prefix je, tak uz je v string_balance
        numof_prefixes+=string_balance.count(desired_balance)
        print(numof_prefixes)
        continue

    for bal in string_balance:
        if (desired_balance-bal)%periodic_addition==0 \
            and ((periodic_addition<0 and bal>=desired_balance) or(periodic_addition>0 and bal<=desired_balance)):
            numof_prefixes+=1
    print(numof_prefixes, end="\n")