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 dp[2000];
char str[2000], tr[2000];
int main() {
  int i, j, k, l, m, n, q;
  scanf("%d%d%d", &n, &m, &q);
  scanf("%s", str + 1);
  scanf("%s", tr + 1);
  for (int i = 1; str[i]; i++) {
    for (j = 1, k = i; tr[j] && str[k]; j++, k++) {
      if (tr[j] != str[k]) break;
    }
    dp[i] += dp[i - 1];
    if (j <= m) continue;
    dp[i]++;
  }
  while (q--) {
    scanf("%d%d", &l, &k);
    if ((k - l + 1) < m) {
      printf("0\n");
      continue;
    }
    int v = k - m + 1;
    int ans = dp[v] - dp[l - 1];
    cout << ans << endl;
  }
}
