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;
struct re {
  int x, y, cs;
};
const long long module = 1000000000 + 7;
const int maxN = 1000 + 10;
int n, m, q, kq[maxN];
string s, t;
void nhap() { cin >> n >> m >> q >> s >> t; }
int kt(int x) {
  for (int i = 1; i <= m - 1; ++i)
    if (s[x + i] != t[i]) return (0);
  return (1);
}
void xuly() {
  for (int i = 0; i <= n - m; ++i) {
    if (i > 0) kq[i] = kq[i - 1];
    if (s[i] == t[0]) kq[i] += kt(i);
  }
}
void xuat() {
  int x, y;
  for (int i = 1; i <= q; ++i) {
    cin >> x >> y;
    x--;
    y--;
    if (y - x + 1 < m)
      cout << "0";
    else if (x == 0)
      cout << kq[y - m + 1];
    else
      cout << kq[y - m + 1] - kq[x - 1];
    cout << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  nhap();
  xuly();
  xuat();
  return 0;
}
