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.
from bisect import bisect_left, bisect_right

n, m, q = map(int, raw_input().split(" "))
s = raw_input()
t = raw_input()

def occurrences(string, sub):
    count = start = 0
    starts = []
    while True:
        start = string.find(sub, start)
        if start >= 0:
            starts.append(start)
            start += 1
        else:
            return starts

tut = occurrences(s, t)
tut = map(lambda x: x+1, tut)


# function to find first index >= x
def lowerIndex(arr, n, x):
    l = 0
    h = n-1
    while (l <= h):
        mid = int((l + h)/2)
        if (arr[mid] >= x):
            h = mid - 1
        else:
            l = mid + 1
    return l


# function to find last index <= x
def upperIndex(arr, n, x):
    l = 0
    h = n-1
    while (l <= h):
        mid = int((l + h)/2)
        if (arr[mid] <= x):
            l = mid + 1
        else:
            h = mid - 1
    return h


# function to count elements within given range
def countInRange(arr, n, x, y):
    # initialize result
    count = 0;
    count = upperIndex(arr, n, y) - lowerIndex(arr, n, x) + 1;
    return count

for i in xrange(q):
    l, r = map(int, raw_input().split(" "))

    if r-l+1 < len(t) or r < l:
        print 0
    else:
        r = r - len(t) + 1
        #print tut
        #print l, r
        print countInRange(tut, len(tut), l, r)

