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:
# Every repeat of the s has balance
# If the balance is positive, the balance is strictly increasing,
# If the balance is negative, the balance is strictly decreasing,
# If the balance is zero, if s has such x, there are INFINITE x's.

def str_balance(s):
    return s.count("0") - s.count("1")

def val(c):
    if c == "1":
        return -1
    return 1

def arr_balance(s):
    l = [0]
    for c in s:
        l.append(l[-1] + val(c))
    return l[1:]
    
INFINITE = -1
def calc_prefix_number(s, n, x):
    l = arr_balance(s)
    steps = l[-1]
    if steps == 0:
        if x in l:
            return INFINITE
        return 0
    cnt = 0
    if x == 0:
        cnt += 1
    for cell in l:
        if (x - cell) % steps == 0 and (x - cell) * steps >= 0:
            cnt += 1
    return cnt

testcases = int(input())
for _ in range(testcases):
    n, x = [int(p) for p in input().split()]
    s = input()
    print(calc_prefix_number(s, n, x))