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:
#######################################################################################################################
#   Author:     BlackFyre
#   Language:   PyPy 3.7
#######################################################################################################################
 
from sys import stdin, stdout, setrecursionlimit
from math import floor, gcd, fabs, factorial, fmod, sqrt, inf, log, log2
from random import seed, randint
from datetime import datetime
from collections import defaultdict as dd, deque
from heapq import merge, heapify, heappop, heappush, nsmallest
from bisect import bisect_left as bl, bisect_right as br, bisect
from collections import defaultdict as dd
mod = pow(10, 9) + 7
mod2 = 998244353
# setrecursionlimit(3000)
 
def inp(): return stdin.readline().strip()
def iinp(): return int(inp())
def out(var, end="\n"): stdout.write(str(var) + "\n")
def outa(*var, end="\n"): stdout.write(' '.join(map(str, var)) + end)
def lmp(): return list(mp())
def mp(): return map(int, inp().split())
def smp(): return map(str, inp().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(m, val) for j in range(n)]
def remadd(x, y): return 1 if x % y else 0
def ceil(a, b): return (a + b - 1) // b
def def_value(): return 0
def def_inf(): return inf
def atoi(s): return ord(s)-ord('a')
def inc(s, n=1): return chr((ord(s)-ord('a')+n)%26+ord('a'))

class XY:
    def __init__(self, x, y):
        self.x = x
        self.y = y
        
def rotate(a,b,c):
    ang = (b.y-a.y)*(c.x-b.x) - (b.x-a.x)*(c.y-b.y)
    if ang == 0: return 0   #0coll, +-clock, --aclock
    else: return ang/abs(ang)
    
   
for _ in range(iinp()):
    n,x = mp()
    s = inp()
    b = s.count('0')-s.count('1')
    
    c0,c1 = 0,0
    bl = l1d(n+1,0)
    for i in range(1,n+1):
        if s[i-1]=="0": c0+=1
        else: c1+=1
        bl[i] = c0-c1
    
    cnt = 0
    if b==0:
        if x in bl: print(-1)
        else: print(0)
    else:
        cnt = 0
        for i in bl[:n]:
            if (x-i)%b == 0 and (x-i)//b >= 0:
                cnt += 1
        print(cnt)