Segment Occurrences(substr函数)

ou are given two strings ss and tt, both consisting only of lowercase Latin letters.

The substring s[l..r]s[l..r] is the string which is obtained by taking characters sl,sl+1,…,srsl,sl+1,…,sr without changing the order.

Each of the occurrences of string aa in a string bb is a position ii (1≤i≤|b|−|a|+11≤i≤|b|−|a|+1) such that b[i..i+|a|−1]=ab[i..i+|a|−1]=a (|a||a| is the length of string aa).

You are asked qq queries: for the ii-th query you are required to calculate the number of occurrences of string tt in a substring s[li..ri]s[li..ri].

Input

The first line contains three integer numbers nn, mm and qq (1≤n,m≤1031≤n,m≤103, 1≤q≤1051≤q≤105) — the length of string ss, the length of string tt and the number of queries, respectively.

The second line is a string ss (|s|=n|s|=n), consisting only of lowercase Latin letters.

The third line is a string tt (|t|=m|t|=m), consisting only of lowercase Latin letters.

Each of the next qq lines contains two integer numbers lili and riri (1≤li≤ri≤n1≤li≤ri≤n) — the arguments for the ii-th query.

Output

Print qq lines — the ii-th line should contain the answer to the ii-th query, that is the number of occurrences of string tt in a substring s[li..ri]s[li..ri].

Examples

input

Copy

10 3 4
codeforces
for
1 3
3 10
5 6
5 7

output

Copy

0
1
0
1

input

Copy

15 2 3
abacabadabacaba
ba
1 15
3 4
2 14

output

Copy

4
0
3

input

Copy

3 5 2
aaa
baaab
1 3
1 1

output

Copy

0
0
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int a[1050];
int main()
{
    int n,m,k,q,l,r,ans;
    cin>>n>>m>>q;
    memset(a,0,sizeof(a));
    string s,t;
    cin>>s>>t;
    for(int i=0;i<=n-m;i++)
    if(s.substr(i,m)==t)
    a[i]=1;
    while(q--)
    {ans=0;
        int l,r;
        cin>>l>>r;
        for(int i=l-1;i<=r-m;i++)
        if(a[i])
        ans++;
        cout<<ans<<endl;
    }
    return 0;
}

猜你喜欢

转载自blog.csdn.net/sdauguanweihong/article/details/81635025