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.
#E48_B

ln = [int(i) for i in input().split(" ")]
n = ln[0]
m = ln[1]
q = ln[2]

s = input()
t = input()

def runAns(s, t):
    ss = 0
    sso = [0] * (len(t) - 1)

    for i in range(0, n - len(t) + 1):
        f = True
        for j in range(0, n - i):
            if j >= len(t):
                break
            if s[i + j] != t[j]:
                f = False
                break
        if f:
            ss += 1
        sso.append(ss)


    return [ss, sso]

o = runAns(s, t)
ss = o[0]
o1 = o[1]
o = runAns("".join(list(s)[::-1]), "".join(list(t)[::-1]))
o2 = o[1][::-1]

for i in range(0, q):
    ln = [int(j) for j in input().split(" ")]
    n1 = ln[0]
    n2 = ln[1]
    print(max(0, o2[n1 - 1] + o1[n2 - 1] - ss))
