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 mi():
	return map(int, input().split())

n,m,q = mi()
s = list(input())
t = list(input())
pre1 = [0]*n
pre2 = [0]*n
if n>=m:
	for i in range(n-m+1):
		if (s[i:i+m]==t):
			pre1[i+m-1] = 1
			pre2[i] = 1
	for i in range(1,n):
		pre1[i]+=pre1[i-1]
		pre2[i]+=pre2[i-1]
pre1.insert(0,0)
pre2.insert(0,0)


while q:
	q-=1
	l,r = mi()
	s1, s2 = 0,0
	#for i in range(l, r+1):
		#s1+=pre1[i]
		#s2+=pre2[i]
	if m>n:
		print (0)
		continue
	if r<m or r-l+1<m or n-l+1<m:
		print (0)
		continue
	s1 = pre1[r]-pre1[l-1+m-1]
	s2 = pre2[r-m+1]-pre2[l-1]
	if s1>0 and s2>0:
		print (min(s1,s2))
	else:
		print (0)
	
