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.
tmp=input().split()
n=int(tmp[0])
m=int(tmp[1])
q=int(tmp[2])
arr1=[]
arr2=[]
tmp=input()
for i in tmp:
	arr1.append(i)
tmp=input()
for i in tmp:
	arr2.append(i)
hash=[0 for i in range(n+10)]
for i in range(n-m+1):
	if(arr2==arr1[i:i+m:1]):
		hash[i+1]=1
for i in range(1,n+2,1):
	hash[i]+=hash[i-1]
for i in range(q):
	tmp=input().split()
	l=int(tmp[0])
	r=int(tmp[1])
	if(r-l+1<m or m>n):
		ans=0
	else:
		ans=hash[r-m+1]-hash[l-1]
	print(ans)