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;
char s1[1005], s2[1005];
int s1len, s2len;
int Next[1005];
void kmp() {
  int i = 0;
  Next[i] = -1;
  int j = -1;
  while (i < s1len) {
    if (j == -1 || s1[i] == s1[j]) {
      i++;
      j++;
      Next[i] = j;
    } else {
      j = Next[j];
    }
  }
}
int main(void) {
  int Q, L, R;
  scanf("%d%d%d", &s2len, &s1len, &Q);
  scanf("%s", s2);
  scanf("%s", s1);
  kmp();
  while (Q--) {
    scanf("%d%d", &L, &R);
    int i = L - 1, j = 0;
    int ans = 0;
    while (i < R) {
      if (j == -1 || s2[i] == s1[j]) {
        i++;
        j++;
      } else {
        j = Next[j];
      }
      if (j == s1len) {
        ans++;
        j = Next[j];
      }
    }
    printf("%d\n", ans);
  }
  return 0;
}
