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

def minp():
	return sys.stdin.readline().strip()

def mint():
	return int(minp())

def mints():
	return map(int, minp().split())

def solve():
	n, x = mints()
	s = minp()
	p = 0
	all = dict()
	m = int(1e9)
	M = int(-1e9)
	for i in range(n):
		all[p] = all.get(p,0) + 1
		m = min(m, p)
		M = max(M, p)
		if s[i] == '0':
			p += 1
		else:
			p -= 1
	#print(m, M, all)
	if p == 0:
		if x in all:
			print(-1)
		else:
			print(0)
		return
	ss = 1
	if p < 0:
		p = -p
		m, M = -M, -m
		ss = -1
		x = -x
	do = max((x - M) // p, 0)
	r = 0
	while x - do * p >= m:
		y = ss*(x - p * do)
		#print(do)
		if y in all:
			r += all[y]
		do += 1
	print(r)

for i in range(mint()):
	solve()