You are given two strings s and t, both consisting only of lowercase Latin letters.

The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.

Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).

You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].

Input

The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.

The second line is a string s (|s| = n), consisting only of lowercase Latin letters.

The third line is a string t (|t| = m), consisting only of lowercase Latin letters.

Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.

Output

Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].

Examples

Input

10 3 4
codeforces
for
1 3
3 10
5 6
5 7


Output

0
1
0
1


Input

15 2 3
abacabadabacaba
ba
1 15
3 4
2 14


Output

4
0
3


Input

3 5 2
aaa
baaab
1 3
1 1


Output

0
0

Note

In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
#include <bits/stdc++.h>
using namespace std;
int n, m, q;
char a[1005], b[1005];
struct qujian {
  int s, e;
};
qujian qq[10005];
int num = 0;
int main() {
  scanf("%d%d%d", &n, &m, &q);
  scanf("%s\n%s", a, b);
  if (n < m) {
    for (int i = 0; i < q; i++) {
      int x, y;
      scanf("%d%d", &x, &y);
      printf("0\n");
    }
  } else {
    for (int i = 0; i <= n - m; i++) {
      if (strncmp((a + i), b, m) == 0) {
        qq[num].s = i;
        qq[num++].e = i + m - 1;
      }
    }
    for (int i = 0; i < q; i++) {
      int x, y;
      scanf("%d%d", &x, &y);
      if (y - x + 1 < m)
        printf("0\n");
      else {
        x = x - 1;
        y = y - 1;
        int flag = 0, tnum = 0;
        for (int i = 0; i < num; i++) {
          if (qq[i].s >= x && qq[i].e <= y) {
            tnum++;
          }
        }
        printf("%d\n", tnum);
      }
    }
  }
  return 0;
}
