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.
def main():
  [n,m,q] = list(map(int, input().split(" ")))
  s = input()
  t = input()
  queries = []
  for _ in range(q):
    queries.append(list(map(int, input().split(" "))))
  #toss the obvious
  if len(s) < len(t):
    for _ in range(q):
      print(0)
  else:
    ref = [0]
    for i in range(len(s)-len(t)+1):
      j = i
      #print(j)
      if s[j:j+len(t)] == t:
        ref.append(ref[-1]+1)
      else:
        ref.append(ref[-1])
    #print(ref)
    for x in queries:
      if x[1]-len(t)+1 < x[0]-1:
        print(0)
      else:
        print(ref[x[1]-len(t)+1]-ref[x[0]-1])

main()