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 long long mod = 1e9 + 7;
const int maxn = 1010;
char a[maxn], b[maxn];
int n, m, q;
vector<int> pos;
void solve() {
  while (q--) {
    int l, r, tr;
    scanf("%d%d", &l, &r);
    tr = r - m + 1;
    if (l > tr)
      puts("0");
    else {
      printf("%d\n",
             upper_bound(pos.begin(), pos.end(), tr) - pos.begin() -
                 (lower_bound(pos.begin(), pos.end(), l) - pos.begin()));
    }
  }
}
int main() {
  cin >> n >> m >> q;
  cin >> a + 1;
  cin >> b + 1;
  for (int i = 1; i <= n; ++i) {
    bool ok = true;
    for (int j = 0; j < m; ++j) {
      if (i + j > n || (a[i + j] != b[j + 1])) {
        ok = false;
        break;
      }
    }
    if (ok) pos.push_back(i);
  }
  solve();
  return 0;
}
