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>
const long long N = 1e6 + 10;
const long long mod = 998244353;
using namespace std;
long long w[1111];
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, m, q;
  cin >> n >> m >> q;
  string a, b;
  cin >> a >> b;
  for (long long i = 0; i < n - m + 1; i++) {
    long long ff = 0;
    if (a[i] == b[0]) {
      for (long long j = i, k = 0; j < a.size() && k < b.size(); j++, k++) {
        if (a[j] != b[k]) {
          ff = 1;
          break;
        }
      }
      if (ff == 0) w[i + 1] = 1;
    }
  }
  for (long long i = 1; i <= 1010; i++) w[i] += w[i - 1];
  while (q--) {
    long long l, r;
    cin >> l >> r;
    r -= m;
    r++;
    if (r < l)
      cout << 0 << endl;
    else
      cout << w[r] - w[l - 1] << endl;
  }
  return 0;
}
