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:


def solve_step3(res, test, balance, keys):
    mi, ma = min(keys), max(keys)
    goal = test[1]
    if balance > 0:
        if goal > ma:
            goal -= balance*((goal-ma)//balance)
            if goal > ma:
                goal -= balance
        sum1 = 0
        while goal >= mi:
            sum1 += res[goal]
            goal -= balance
        print(sum1)
    else:
        if goal < mi:
            goal -= (balance*((goal-mi)//balance))
            if goal < mi:
                goal -= balance#proceed to next iteration of s
        sum1 = 0
        while goal <= ma:
            sum1 += res[goal]
            goal -= balance
        print(sum1)





def solve_step2(res, test, balance):
    goal = test[1]
    keys = [i for i in res.keys() if res[i] > 0]
    if balance == 0:
        if goal in keys:
            print("-1")
        else:
            print("0")
        return
    else:
        solve_step3(res, test, balance, keys)


def solve_step1(test):
    string = test[0]
    length = len(string)
    possibles = {i: (False, 0) for i in range(-1*length, length+1)}
    sum1 = 0
    for i in range(length):
        possibles[sum1] = (True, possibles[sum1][1] + 1)
        num = int(string[i])
        sum1 += pow(-1, num)
    res = {i: possibles[i][1] for i in possibles}
    solve_step2(res, test, sum1)



def handle_tests():
    number_of_tests = int(input())
    tests = []
    for i in range(number_of_tests):
        n, x = input().split()
        string = input()
        tests.append((string, int(x)))
    for test in tests:
        solve_step1(test)

handle_tests()