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:
import sys

t = int(sys.stdin.readline())

for _ in range(t):
	n,m = list(map(int, sys.stdin.readline().split()))
	re = 0
	if m == 0:
		re += 1
	s = sys.stdin.readline().split()[0]
	n_sum = []
	for i, c in enumerate(s):
		if i == 0:
			n_sum.append(1 if c == '0' else -1)
		else:
			n_sum.append(n_sum[i-1] + (1 if c == '0' else -1))
	cycle = n_sum[-1:][0]
	if cycle == 0:
		flag = False
		for i in n_sum:
			if i == m:
				flag = True
				break
		if flag:
			print(-1)
		else:
			print(0)
		continue
	for i in n_sum:
		if((m - i)%cycle == 0 and int((m - i)/cycle)>=0):
			re += 1
	print(re)