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 fx[] = {-1, 1, 0, 0};
int fy[] = {0, 0, -1, 1};
int dx[] = {1, 1, 0, -1, -1, -1, 0, 1};
int dy[] = {0, 1, 1, 1, 0, -1, -1, -1};
int kx[] = {1, 1, -1, -1, 2, 2, -2, -2};
int ky[] = {2, -2, 2, -2, 1, -1, 1, -1};
bool cmp(const pair<int, int> &a, const pair<int, int> &b) {
  return a.first < b.first;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m, q;
  cin >> n >> m >> q;
  string s, t;
  cin >> s >> t;
  s = '?' + s;
  int ara[1010], sum[1010];
  memset(ara, 0, sizeof(ara)), memset(sum, 0, sizeof(sum));
  for (int i = 1; i <= n - m + 1; i++) {
    ara[i] = 1;
    for (int j = 0; j < t.size(); j++) {
      if (s[i + j] != t[j]) {
        ara[i] = 0;
        break;
      }
    }
    sum[i] = sum[i - 1] + ara[i];
  }
  for (int i = 1; i <= q; i++) {
    int l, r;
    cin >> l >> r;
    if (r - l + 1 < m) {
      cout << "0\n";
    } else {
      cout << sum[r - m + 1] - sum[l - 1] << "\n";
    }
  }
  return 0;
}
