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):
	n_x = list(map(int, input().split(" ")))
	n, x = n_x
	pre = input()
	
	count = 0
	sett = set()
	diff = 0
	for i in pre:
		if i == "0":
			diff += 1
			sett.add(diff)
		else:
			diff -= 1
			sett.add(diff)

	if diff != 0 and x == 0:
		count += 1

	if diff == 0:
		if x in sett:
			print(-1)
		else:
			print(0)
	else:
		new_diff = 0
		for i in pre:
			if i == "0":
				new_diff += 1
			else:
				new_diff -= 1

			if diff < 0 and new_diff >= x and abs(new_diff-x) % diff == 0:
					count += 1
			elif diff > 0 and new_diff <= x and abs(new_diff-x) % diff == 0:
					count += 1
		print(count)