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.
# Python program for KMP Algorithm 
def KMPSearch(pat, txt): 
	ans = []
	M = len(pat) 
	N = len(txt) 

	# create lps[] that will hold the longest prefix suffix 
	# values for pattern 
	lps = [0]*M 
	j = 0 # index for pat[] 

	# Preprocess the pattern (calculate lps[] array) 
	computeLPSArray(pat, M, lps) 

	i = 0 # index for txt[] 
	while i < N: 
		if pat[j] == txt[i]: 
			i += 1
			j += 1

		if j == M: 
			ans.append(i-j)
			j = lps[j-1] 

		# mismatch after j matches 
		elif i < N and pat[j] != txt[i]: 
			# Do not match lps[0..lps[j-1]] characters, 
			# they will match anyway 
			if j != 0: 
				j = lps[j-1] 
			else: 
				i += 1
	return ans

def computeLPSArray(pat, M, lps): 
	len = 0 # length of the previous longest prefix suffix 

	lps[0] # lps[0] is always 0 
	i = 1

	# the loop calculates lps[i] for i = 1 to M-1 
	while i < M: 
		if pat[i]== pat[len]: 
			len += 1
			lps[i] = len
			i += 1
		else: 
			# This is tricky. Consider the example. 
			# AAACAAAA and i = 7. The idea is similar 
			# to search step. 
			if len != 0: 
				len = lps[len-1] 

				# Also, note that we do not increment i here 
			else: 
				lps[i] = 0
				i += 1

n, m, q = map(int, input().split())
s = input().strip()
t = input().strip()
pos = KMPSearch(t, s)

for i in range(q):
	l, r = map(int, input().split())
	l -= 1
	ans = 0
	for a in pos:
		if(a >= l and a+m <=r): ans+=1
	print(ans)
