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;
const int N = 1e3 + 5;
int n, m, q, dp[N][N], p[N + N], used[N + N];
string s, t;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> q >> s >> t;
  string x = t + "#" + s;
  for (int i = 1; i < (int)x.size(); i++) {
    int j = p[i - 1];
    while (j > 0 && x[i] != x[j]) {
      j = p[j - 1];
    }
    if (x[i] == x[j]) {
      j++;
    }
    p[i] = j;
    if (p[i] == m) {
      used[i - m] = true;
    }
  }
  s = '#' + s;
  t = '%' + t;
  for (int i = 1; i <= n; i++) {
    for (int j = i; j <= n; j++) {
      dp[i][j] = dp[i][j - 1] + (j - m + 1 >= i && used[j]);
    }
  }
  while (q--) {
    int l, r;
    cin >> l >> r;
    cout << dp[l][r] << "\n";
  }
}
