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 count_overlapping_substrings(haystack, needle):
    count = 0
    i = -1
    while True:
        i = haystack.find(needle, i + 1)
        if i == -1:
            return count
        left.append(i)
        right.append(i + len(needle) - 1)
        count += 1


def bin_search(lst, x):
    lower_bound = 0
    upper_bound = len(lst) - 1
    while lower_bound != upper_bound:
        compared_value = (lower_bound + upper_bound) // 2  # Целочисленный тип в Python имеет неограниченную длину
        if x == lst[compared_value]:
            return compared_value
        elif x < lst[compared_value]:
            upper_bound = compared_value
        else:
            lower_bound = compared_value + 1
    return lower_bound


n, m, q = map(int, input().split())
s1 = input()
s2 = input()
right = []
left = []

count_overlapping_substrings(s1, s2)

for i in range(q):
    l, r = map(int, input().split())
    if right == []:
        print(0)
        continue
    a = bin_search(left, l - 1)
    b = bin_search(right, r - 1)
    if left[a] < l - 1:
        a += 1
    if right[b] > r - 1:
        b -= 1
    if (a > b):
        print(0)
    else:
        print(b - a + 1)
