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(input())
for e in range(t):
	
	n, x = map(int, input().split())
	dp = [0 for i in range(n + 1)]
	s = input()
	cur = 0
	for i in range(n):
		if(s[i] == '0'):
			cur += 1	
		else:
			cur -= 1
		dp[i + 1] = cur
	# print(dp)

	if(dp[-1] == 0):
		if(min(dp) <= x <= max(dp)):
			print(-1)
		else:
			print(0)
		continue
	cnt = 0
	for i in range(n):
		diff = x-dp[i]
		# print(i, diff)
		if((diff)*dp[-1] >= 0 and diff%dp[-1] == 0):
			cnt += 1
	print(cnt)