Segment Occurrences（codeforces 1016B）

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本文介绍了一种解决字符串匹配问题的方法，具体来说，如何在一个较长的字符串中查找特定子串出现的位置和次数，通过预处理和查询操作实现高效求解。

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Description

You 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 ttand 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].

Sample Input

Input

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

Output

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

Hint

In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.

题解：

给出两个长为n和m的字符串s和t，给出q次查询，每次给出长度l和r表示查询在字符串s中有多少个t的子串。用substr不断取出上为m的子串，并记录下来，查询的时候直接枚举查询的区间。不了解substr的小伙伴点击这里。某个同学用的find查找，每找到一次记录位置，并将首字母修改，这样可以查找出所有的子串。

代码如下：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<iomanip>
#include<map>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<utility>
#include<list>
#include<algorithm>
#include <ctime>
#define max(a,b)   (a>b?a:b)
#define min(a,b)   (a<b?a:b)
#define swap(a,b)  (a=a+b,b=a-b,a=a-b)
#define memset(a,v)  memset(a,v,sizeof(a))
#define X (sqrt(5)+1)/2.0
#define maxn 320007
#define N 200005
#define INF 0x3f3f3f3f
#define PI acos(-1)
#define lowbit(x) (x&(-x))
#define read(x) scanf("%d",&x)
#define put(x) printf("%d\n",x)
#define memset(x,y) memset(x,y,sizeof(x))
#define Debug(x) cout<<x<<" "<<endl
#define lson i << 1,l,m
#define rson i << 1 | 1,m + 1,r
#define mod 1000000009
#define e  2.718281828459045
#define eps 1.0e-8
#define ll long long
using namespace std;



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