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 cnt0,qcnt1,q, where cnt0,q is the number of occurrences of 0 in q, and cnt1,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 ≤ 105, −109 ≤ x ≤ 109) — the length of string s and the desired balance, respectively.

The second line contains the binary string s (|s| = n, si ∈ {0,1}).

It’s guaranteed that the total sum of n doesn’t exceed 105.

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.

Problem solving:

这场cf全是字符串,直接自闭。

题目意思是给你一个01字符串,然后这个字符串可以无限叠加。例如010010可以变成010010 010010010010 010010 010010 010010 010010(串中没有空格) … 。然后找这个无限字符串的前缀,前缀0的个数减去1的个数等于x。问有多少种符合的前缀。

因为x有正负之分,需要判断的太多了,然后就把自己绕进去了。看了别人的代码后才知道可以先把负数的 x 变成正数,然后把原串01对换,这样就只需要判断正数的x了。

把原串 cnt0,qcnt1,q 记为 num

因为串是无穷的,所以会出现有无穷多个解的情况。当原串 num 等于 0 ,原串前缀的cnt0,qcnt1,q等于x 时,答案就有无穷多个。输出 -1

num 不为 0

我用一个数组a记录原串中前缀当前位置的 cnt0,icnt1,i ,如下表所示:

s = 010010 , x = 10

然后遍历这个数组,当 ( x - a[i] ) % num == 0 && ( x - a[i] ) / num >= 0 ) 时 答案ans+1

如表中 下标为2时 ,a[1] = 0,( x - a[i] ) % num == 0( x - a[i] ) / num = 5 。说明前缀包含了五个原串中,再加上第六个原串的前两个就可以满足 cnt0,qcnt1,q = 10 当前位置为 5 * 6 + 232

当下标为4时, ( x - a[i] ) / num = 4 ,我们可以得到在位置 4 * 6 + 428 处找到解。

当下标为6时, ( x - a[i] ) / num = 4 ,我们可以得到在位置 4 * 6 + 630 处找到解。

符合样例一。

AC代码:

#include<bits/stdc++.h>
#define fi        first
#define se        second
#define PI        acos(-1)
#define LC(a)     ((a<<1))
#define RC(a)     ((a<<1)+1)
#define MID(a,b)  ((a+b)>>1)
#define mem(a, b) memset(a, b, sizeof(a))
#define IOS()     std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
using namespace std;
typedef long long ll;
typedef unsigned long long ULL;
typedef pair<int,int> PII;
typedef pair<ll,ll>   PLL;
typedef pair<ll,int>  PLI;
const int INF = 0X3F3F3F3F;
const int MIN = -(1<<30);
const ll N = 1e6+7;
ll a[N];
int main()
{
    IOS();
    int t;
    cin>>t;
    while(t--)
    {
        ll n,x;
        string s;
        cin>>n>>x>>s;
        if(x<0)
        {
            x=-x;
            for(int i=0; i<n; i++)
                s[i]=s[i]=='0'?'1':'0';
        }
//        cout<<s<<endl;
        ll num=0;
        for(int i=0; i<s.size(); i++)
        {
            if(s[i]=='0') num++;
            else num--;
            a[i]=num;
        }
        ll ans=0;
        int flag = 0;

        if(num==0)
        {
            for(int i=0; i<n; i++)
            {
                if(a[i]==x) flag=1;
            }
        }
        else
        {
            for(int i=0; i<n; i++)
            {
                if((x-a[i])%num==0 && (x-a[i])/num>=0) ans++;
            }
        }
        if(x==0) ans++;
        if(flag) cout<<-1<<endl;
        else cout<<ans<<endl;

    }
    return 0;
}

补题过程中我输出 -1 用了 puts,导致了不断地 Wrong answer on test 4 ,百度后才知道 关闭同步流 (std::ios::sync_with_stdio(false) )之后,不能再使用 scanf 、printf、gets、puts …。昨天有一道题换行用了endl也导致了wa,改用 ‘\n’ 就 ac,难受了我好长时间。我决定以后慢慢地放弃cin、cout。 虽然它很香。


一个好奇的人