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.
#!/usr/bin/python3

def calc(l, r, layers, li):
    if r - l == 1:
        return layers[li][l]

    xl = l
    if l % 2 == 1:
        xl -= 1
    xr = r
    if r % 2 == 1:
        xr += 1

    ans = calc(xl // 2, xr // 2, layers, li + 1)
    if xl != l:
        ans -= layers[li][xl]
    if xr != r:
        ans -= layers[li][xr - 1]
    return ans


def solve(N, M, Q, S, T, A):
    base = [0] * N
    i = 0
    while True:
        try:
            i = S.index(T, i)
            base[i] = 1
            i += 1
        except ValueError:
            break

    layers = [base]
    while len(layers[-1]) > 1:
        prev = layers[-1]
        if len(prev) % 2 == 1:
            prev.append(0)

        layer = []
        for i in range(len(prev) // 2):
            layer.append(prev[2 * i] + prev[2 * i + 1])

        layers.append(layer)

    ans = []
    for (l, r) in A:
        l -= 1

        r -= M - 1
        if r <= l:
            ans.append(0)
            continue

        ans.append(calc(l, r, layers, 0))

    return ans


def main():
    N, M, Q = [int(e) for e in input().split(' ')]
    S = input()
    T = input()
    A = [map(int, input().split(' ')) for _ in range(Q)]

    assert len(S) == N
    assert len(T) == M

    print(*solve(N, M, Q, S, T, A), sep='\n')


if __name__ == '__main__':
    main()
