Problem: 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 S of m "01-strings".

A "01-string" is a string that contains n characters "0" and "1". For example, if n=4, strings "0110", "0000", and "1110" are "01-strings", but "00110" (there are 5 characters, not 4) and "zero" (unallowed characters) are not.

Note that the multiset S can contain equal elements.

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

Mrs. Kasoura and Mr. Kasoura think that if s_i = t_i (1≤ i≤ n) then the "Wu" value of the character pair equals to w_i, otherwise 0. 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 n.

For example, if w=[4, 5, 3, 6], "Wu" of ("1001", "1100") is 7 because these strings have equal characters only on the first and third positions, so w_1+w_3=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 S such that the "Wu" value of the pair is not greater than k.

Input

The first line contains three integers n, m, and q (1≤ n≤ 12, 1≤ q, m≤ 5⋅ 10^5) — the length of the "01-strings", the size of the multiset S, and the number of queries.

The second line contains n integers w_1, w_2, …, w_n (0 ≤ w_i ≤ 100) — the value of the i-th caracter.

Each of the next m lines contains the "01-string" s of length n — the string in the multiset S.

Each of the next q lines contains the "01-string" t of length n and integer k (0≤ k≤ 100) — the query.

Output

For each query, print the answer for this query.

Examples

Input

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


Output

2
4
2
3
4


Input

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


Output

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 2, not 1.

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

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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int M = 1000000000 + 7;
const int N = 500000 + 10;
const int L = 4000 + 100;
const int K = 100 + 10;
struct re {
  int x, y;
};
int n, m, q, w[N], cnt[L], f[L][K], sum[L][K], g[L][L];
char s[20];
int cal(int x, int y, int n) {
  if (n == 0) return 0;
  return cal(x / 2, y / 2, n - 1) + w[n] * (x % 2 == y % 2);
}
int main() {
  scanf("%d %d %d", &n, &m, &q);
  for (int i = (1); (i <= (n)); ++i) scanf("%d", &w[i]);
  for (int i = (1); (i <= (m)); ++i) {
    scanf("%s", s);
    int tmp = 0;
    for (int j = (0); (j <= (n - 1)); ++j) tmp = tmp * 2 + s[j] - '0';
    cnt[tmp]++;
  }
  int k = -1;
  for (int i = (0); (i <= ((1 << n) - 1)); ++i) {
    if (1 & (i >> (k + 1))) k++;
    for (int j = (0); (j <= ((1 << n) - 1)); ++j) {
      if (i == 0)
        g[i][j] = cal(i, j, n);
      else
        g[i][j] = g[i - (1 << k)][j] + ((1 & (j >> k)) ? w[n - k] : -w[n - k]);
      if (g[i][j] <= 100) f[i][g[i][j]] += cnt[j];
    }
  }
  for (int i = (0); (i <= ((1 << n) - 1)); ++i) {
    sum[i][0] = f[i][0];
    for (int j = (1); (j <= (100)); ++j) sum[i][j] = sum[i][j - 1] + f[i][j];
  }
  while (q--) {
    int k;
    scanf("%s %d", s, &k);
    int tmp = 0;
    for (int j = (0); (j <= (n - 1)); ++j) tmp = tmp * 2 + s[j] - '0';
    printf("%d\n", sum[tmp][k]);
  }
}