CodeForces - 1017D Round #502 D. The Wu（状压预处理）

D. The Wu

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Childan is making up a legendary story and trying to sell his forgery — a necklace with a strong sense of "Wu" to the Kasouras. But Mr. Kasoura is challenging the truth of Childan's story. So he is going to ask a few questions about Childan's so-called "personal treasure" necklace.

This "personal treasure" is a multiset 

A "01-string" is a string that contains 

Note that the multiset S can contain equal elements.

Frequently, Mr. Kasoura will provide a "01-string" k.

Mrs. Kasoura and Mr. Kasoura think that if n.

For example, if w1+w3=4+3=7.

You need to help Childan to answer Mr. Kasoura's queries. That is to find the number of strings in the multiset k.

Input

The first line contains three integers S, and the number of queries.

The second line contains i-th caracter.

Each of the next S.

Each of the next 0≤k≤100) — the query.

Output

For each query, print the answer for this query.

Examples

input

Copy

2 4 5
40 20
01
01
10
11
00 20
00 40
11 20
11 40
11 60

output

Copy

2
4
2
3
4

input

Copy

1 2 4
100
0
1
0 0
0 100
1 0
1 100

output

Copy

1
2
1
2

Note

In the first example, we can get:

"Wu" of ("01", "00") is 40.

"Wu" of ("10", "00") is 20.

"Wu" of ("11", "00") is 0.

"Wu" of ("01", "11") is 20.

"Wu" of ("10", "11") is 40.

"Wu" of ("11", "11") is 60.

In the first query, pairs ("11", "00") and ("10", "00") satisfy the condition since their "Wu" is not greater than 20.

In the second query, all strings satisfy the condition.

In the third query, pairs ("01", "11") and ("01", "11") satisfy the condition. Note that since there are two "01" strings in the multiset, the answer is 1.

In the fourth query, since 

In the fifth query, since 

 

 

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#define MAX 500005
#define INF 0x3f3f3f3f
#define MOD 1000000007
using namespace std;
typedef long long ll;

int a[MAX];
char s[MAX][20];
char ss[20];
ll dp[(1<<12)+5][105];
int b[(1<<12)+5];

int main()
{
    int n,m,q,x,i,j,k;
    scanf("%d%d%d",&n,&m,&q);
    for(i=0;i<n;i++){
        scanf("%d",&a[i]);
    }
    for(i=0;i<m;i++){
        scanf(" %s",s[i]);
    }
    for(i=0;i<m;i++){
        ll S=0;
        for(j=0;j<n;j++){
            if(s[i][j]=='1') S|=1<<j;
        }
        b[S]++;
    }
    for(i=0;i<(1<<n);i++){
        for(j=0;j<(1<<n);j++){
            if(!b[j]) continue;
            int sum=0;
            for(k=0;k<n;k++){
                if((i&(1<<k))==(j&(1<<k))) sum+=a[k];
            }
            if(sum>100) continue;
            dp[i][sum]+=b[j];
        }
    }
    
    while(q--){
        scanf(" %s %d",ss,&x);
        ll S=0;
        for(i=0;i<n;i++){
            if(ss[i]=='1') S|=1<<i;
        }
        ll ans=0;
        for(i=0;i<=x;i++){
            ans+=dp[S][i];
        }
        printf("%I64d\n",ans);
    }
    return 0;
}