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.
n,m,q=map(int,input().split())
s=input()
s1=input()
st=[]
en=[]
l=[]
for i in range(n):
  st.append(0)
  en.append(0)
for i in range(n):
  if s[i:i+len(s1)] ==s1:
    l.append(1)
    st[i]=1
    en[i+len(s1)-1]=1
  else:
    l.append(0)
sums=0
sume=0
sa=[]
se=[]
a=[]

for i in range(n):
  sums=sums+st[i]
  sume=sume+en[i]
  sa.append(sums)
  se.append(sume)
#print(sa,se)

for i in range(q):
  a1,b1=map(int,input().split())
  #print(se[b1-1],sa[a1-2],b1-1)
  if(a1==1):
    print(se[b1-1]-0)
  else:
    if(se[b1-1]-sa[a1-2]<0):
      print(0)
    else:
      print(se[b1-1]-sa[a1-2])


