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 pref_func(s):
    n = len(s)
    f = list()
    f.append(0)
    for i in range(1,n):
        j = f[i-1]
        while j>0 and s[i] != s[j]:
            j = f[j-1]
        if s[i] == s[j]:
            j += 1
        f.append(j)
    return f


n,m,q = list(map(int,input().split(' ')))
s = input()
t = input()
f = pref_func(t+'$'+s)[m+1:]

y = 0
suf = list()
for i in range(n-1,-1,-1):
    suf.append(y)
    if f[i] == m:
        y += 1
suf.reverse()

for i in range(n):
    if f[i] == m:
        f[i] = 0
        f[i-m+1] = 1
    else:
        f[i] = 0
pre = list()
y = 0
for i in range(n):    
    pre.append(y)
    if f[i] == 1:
        y += 1

mmm = f.count(1)


for i in range(q):
    l,r = list(map(int,input().split(' ')))
    print(max(mmm - pre[l-1] - suf[r-1],0))

