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 mxn = 1e3 + 5, seed = 131, seed2 = 31, MOD = 1e9 + 9;
int n, m, q;
string s, t;
long long p[mxn], pp[mxn], a[mxn], aa[mxn], b[mxn], bb[mxn];
inline long long strHash(int l, int r) {
  return (long long)((1LL * a[r] - 1LL * a[l - 1] * p[r - l + 1] % MOD) + MOD) %
         MOD;
}
int main() {
  cin >> n >> m >> q >> s >> t;
  p[0] = 1;
  for (int i = 1; i <= max(n, m); i++)
    p[i] = ((p[i - 1] * seed % MOD) + MOD) % MOD;
  for (int i = 1; i <= n; i++)
    a[i] = ((a[i - 1] * seed + s[i - 1]) % MOD + MOD) % MOD;
  for (int i = 1; i <= m; i++)
    b[i] = ((b[i - 1] * seed + t[i - 1]) % MOD + MOD) % MOD;
  for (int i = 1, L, R, ans; i <= q; i++) {
    cin >> L >> R;
    ans = 0;
    for (int j = L; j + m - 1 <= R; j++) {
      if (strHash(j, j + m - 1) == b[m]) ans++;
    }
    cout << ans << "\n";
  }
}
