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 main() {
  long long n, m, q, i, j, l, r, a[5000];
  string s, t;
  cin >> n >> m >> q;
  cin >> s >> t;
  if (m > n) goto label;
  if (s.compare(0, m, t) == 0) {
    a[0] = 1;
  } else
    a[0] = 0;
  for (j = 1; j <= n - m + 1; j++) {
    if (s.compare(j, m, t) == 0) {
      a[j] = a[j - 1] + 1;
    } else {
      a[j] = a[j - 1];
    }
  }
  for (j = n - m + 2; j < n; j++) {
    a[j] = a[n - m + 1];
  }
label:
  for (i = 0; i < q; i++) {
    cin >> l >> r;
    l--;
    r--;
    if (r - l < m - 1 || m > n)
      cout << l - l << endl;
    else if (l == 0)
      cout << a[r - m + 1] << endl;
    else
      cout << a[r - m + 1] - a[l - 1] << endl;
  }
  return 0;
}
