1295_B. Infinite Prefixes
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.
```python
import math
tt=int(input())
for i in range(tt):
    n,x=map(int,input().split())
    s=input()
    a=1
    b={}
    def bind(x):
        return -((2*x)-1)
    if x<0:
        a=-1
        x=-x
    sbal=0
    for i in s:
        sbal+=(bind(int(i))*a)
        b[sbal]=b.get(sbal,0)+1
    ans=0
    if sbal==0:
        if b.get(x,0) != 0:
            print(-1)
        else:
            print(0)
        continue
    elif sbal>0:
        add = x - max(b)
        if add > 0:
            x-=math.ceil(add/sbal)*sbal
        for i in range(x,min(b)-sbal,-sbal):
            ans+=b.get(i,0)
    else:
        dec = min(b) - x
        if dec > 0:
            x+=math.ceil(dec/sbal)*sbal
        for i in range(x,max(b)-sbal,-sbal):
            ans+=b.get(i,0)
    if x==0:
        ans+=1
    print(ans)

```