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 SS of mm "01-strings".

A "01-string" is a string that contains nn characters "0" and "1". For example, if n=4n=4, strings "0110", "0000", and "1110" are "01-strings", but "00110" (there are 55 characters, not 44) and "zero" (unallowed characters) are not.

Note that the multiset SS can contain equal elements.

Frequently, Mr. Kasoura will provide a "01-string" tt and ask Childan how many strings ss are in the multiset SS such that the "Wu" value of the pair (s,t)(s,t) is not greater than kk.

Mrs. Kasoura and Mr. Kasoura think that if si=tisi=ti (1≤i≤n1≤i≤n) then the "Wu" value of the character pair equals to wiwi, otherwise 00. The "Wu" value of the "01-string" pair is the sum of the "Wu" values of every character pair. Note that the length of every "01-string" is equal to nn.

For example, if w=[4,5,3,6]w=[4,5,3,6], "Wu" of ("1001", "1100") is 77 because these strings have equal characters only on the first and third positions, so w1+w3=4+3=7w1+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 SS such that the "Wu" value of the pair is not greater than kk.

Input

The first line contains three integers nn, mm, and qq (1≤n≤121≤n≤12, 1≤q,m≤5⋅1051≤q,m≤5⋅105) — the length of the "01-strings", the size of the multiset SS, and the number of queries.

The second line contains nn integers w1,w2,…,wnw1,w2,…,wn (0≤wi≤1000≤wi≤100) — the value of the ii-th caracter.

Each of the next mm lines contains the "01-string" ss of length nn — the string in the multiset SS.

Each of the next qq lines contains the "01-string" tt of length nn and integer kk (0≤k≤1000≤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 4040.

"Wu" of ("10", "00") is 2020.

"Wu" of ("11", "00") is 00.

"Wu" of ("01", "11") is 2020.

"Wu" of ("10", "11") is 4040.

"Wu" of ("11", "11") is 6060.

In the first query, pairs ("11", "00") and ("10", "00") satisfy the condition since their "Wu" is not greater than 2020.

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 22, not 11.

In the fourth query, since kk was increased, pair ("10", "11") satisfies the condition too.

In the fifth query, since kk was increased, pair ("11", "11") satisfies the condition too.

#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][];
char ss[];
ll dp[(<<)+][];
int b[(<<)+];

int main()
{
    int n,m,q,x,i,j,k;
    scanf("%d%d%d",&n,&m,&q);
    for(i=;i<n;i++){
        scanf("%d",&a[i]);
    }
    for(i=;i<m;i++){
        scanf(" %s",s[i]);
    }
    for(i=;i<m;i++){
        ll S=;
        for(j=;j<n;j++){
            if(s[i][j]=='') S|=<<j;
        }
        b[S]++;
    }
    for(i=;i<(<<n);i++){
        for(j=;j<(<<n);j++){
            if(!b[j]) continue;
            int sum=;
            for(k=;k<n;k++){
                if((i&(<<k))==(j&(<<k))) sum+=a[k];
            }
            if(sum>) continue;
            dp[i][sum]+=b[j];
        }
    }

    while(q--){
        scanf(" %s %d",ss,&x);
        ll S=;
        for(i=;i<n;i++){
            if(ss[i]=='') S|=<<i;
        }
        ll ans=;
        for(i=;i<=x;i++){
            ans+=dp[S][i];
        }
        printf("%I64d\n",ans);
    }
    return ;
}