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.
while True:
	try:
		def failur(p, m):
			failr = [0]*(m+2)
			failr[0]= failr[1] = 0
			i = 2
			while i <= m:
				j = failr[i-1]
				while True:
					if p[i-1] == p[j]:
						failr[i] = j+1
						break
					if j == 0:
						failr[i] = j
						break
					j = failr[j]
				i += 1
			return failr
				
		def matches(t, p, f):
			n = len(t)
			m = len(p)
			c = [0]*(n+2)
			ti = pi = 0
			while True:
				if ti ==  n:
					break
				if t[ti] == p[pi]:
					pi += 1
					ti += 1
					if pi == m:
						pi = f[m]
						c[(ti-m)] += 1
				else:
					if pi == 0:
						 ti += 1
					else:
						pi = f[pi]
			return c
						
			
			
		def solution(n, m, q):
			qury = []
			test = input()
			patt = input()
			for i in range(q):
				l, r = map(int, input().split())
				l -= 1;r -= 1
				qury.append([l, r])
			f = failur(patt, m)
			matc = matches(test, patt, f)
		
			for i in range(q):
				a, b= qury[i][0],  qury[i][1]
				rng = (b -m) +1
				cunt = 0
				for j in range(a, rng+1, 1):
					#print(j, " ", i)
					if matc[j] > 0:
						cunt += 1
				print(cunt)
			#print(matc)
			
			
		def read():
			n, m, q = map(int, input().split())
			solution(n, m, q)
		if __name__  == "__main__":
			read()
	except EOFError:
		break