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.
#list_int 並べて出力 print (' '.join(map(str,ans_li)))
#list_str 並べて出力 print (' '.join(list))

# 2進数 format(10, 'b') # '1010'

''' 二次元配列を一列ずつ
for i in ans:
	print(*i)
'''
''' heapq
queue = []
heapq.heapify(queue) #heapqの作成
heapq.heappush(queue,num) #numのpush(値の追加)
pop = heapq.heappop(queue) #numのpop(最小値の出力)
pop = heapq.heappushpop(queue,num) #push -> pop

'''

from collections import defaultdict
import sys,heapq,bisect,math,itertools,string,queue,datetime
sys.setrecursionlimit(10**8)
INF = float('inf')
mod = 10**9+7
eps = 10**-7
AtoZ = [chr(i) for i in range(65,65+26)]
atoz = [chr(i) for i in range(97,97+26)]

def inpl(): return list(map(int, input().split()))
def inpl_str(): return list(input().split())

N,M,q = inpl()
S = input()
T = input()
azu = [0]*(N-M+1)
qq = []

for i in range(q):
	qq.append(inpl())

if M > N:
	for i in range(q):
		print(0)
	sys.exit()

for i in range(0,N-M+1):
	if S[i:i+M] == T:
		azu[i] = 1

SUM = 0
nyan = [0]
for az in azu:
	SUM += az
	nyan.append(SUM)


for l,r in qq:
	l -= 1
	if r - l < M:
		print(0)
	else:
		print(nyan[r-M+1] - nyan[l])
