1016_B. Segment Occurrences
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
ch=input()
l=list(map(int,ch.split()))
m=l[0]
n=l[1]
p=l[2]
s=input()
t=input()
l1=[]

for i in range(p):
    ch=input()
    l1.append(ch)
res=0
l2=[0]
for i in range (len(t)-1):
    l2.append(0)
for i in range ((len(t)-1),len(s)):
    
    if s[(i-len(t)+1):(i+1)]==t:
                res+=1           
    l2.append(res)
    
for i in range(p):
    
    
    ch=l1[i]
    l=list(map(int,ch.split()))
    r=l[0]
    k=l[1]
    if (k-r+1)>=(len(t)):
        print(l2[k]-l2[(r-2+len(t))])
    else:
        print(0)
```