#pragma GCC optimize ("O3")
#include <iostream>
#include <iomanip>
#include <istream>
#include <ostream>
#include <sstream>
#include <iterator>
#include <vector>
#include <algorithm>
#include <queue>
#include <deque>
#include <list>
#include <stack>
#include <map>
#include <unordered_map>
#include <set>
#include <bitset>
#include <utility>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <string>
#include <ctime>
#include <cctype>
#include <cstdlib>
#define IINF 10e8
#define INF 1<<30
#define MOD 1000000007
#define mod 1000000007
#define REP(i, a, n) for (ll i = a; i < (ll)(n); i++)
#define REPE(i, a, n) for (ll i = a; i <= (ll)(n); i++)
#define Endl endl
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define mt make_tuple
#define eb emplace_back
#define mmax(x,y)(x>y?x:y)
#define mmin(x,y)(x<y?x:y)
#define chmax(x,y) x=mmax(x,y)
#define chmin(x,y) x=mmin(x,y)
#define all(x) (x).begin(),(x).end()
#define siz(x) (ll)(x).size()
#define PI acos(-1.0)
using namespace std;
typedef long long int ll;
typedef unsigned long long int ull;
typedef long double ld;
typedef pair<int,int>Pin;
typedef pair<ll,ll>Pll;
template<class T> using V=vector<T>;
long long GCD(long long a, long long b) {return b?GCD(b,a%b):a;}
long long LCM(long long a, long long b) {return a/GCD(a,b)*b;}
int dx[4]={-1,0,1,0};
int dy[4]={0,-1,0,1};
int ddx[8]={-1,0,1,0,1,1,-1,-1};
int ddy[8]={0,-1,0,1,1,-1,1,-1};
ll cmp(pair<ll,ll>a,pair<ll,ll> b){
        if(a.se!=b.se)
        return a.se<b.se;
        else
        return a.fi<b.fi;
}
//----------------------------------------------------------------------
int N,M,T,P,Q;
char field[1010][1010];
char pattern[1010][1010];
ll hash_[1010][1010],tmp[1010][1010];

void compute_hash(char a[1010][1010],int n,int m){
    const ull B1=9973;
    const ull B2=100000007;
    ull t1=1;
    for(int j=0;j<Q;j++)t1*=B1;
    for(int i=0;i<n;i++){
        ull e=0;
        for(int j=0;j<Q;j++)e=e*B1+a[i][j];
        for(int j=0;j+Q<=m;j++){
            tmp[i][j]=e;
            if(j+Q<m)e=e*B1-t1*a[i][j]+a[i][j+Q];
        }
    }
    ull t2=1;
    for(int i=0;i<P;i++)t2*=B2;
    for(int j=0;j+Q<=m;j++){
        ull e=0;
        for(int i=0;i<P;i++)e=e*B2+tmp[i][j];
        for(int i=0;i+P<=n;i++){
            hash_[i][j]=e;
            if(i+P<n)e=e*B2-t2*tmp[i][j]+tmp[i+P][j];
        }
    }
}
//----------------------------------------------------------------------
int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    //-------------------------------  
    //ll begin_time=clock();
    //-------------------------------
    cin>>N>>M;
    for(int i=0;i<N;i++)for(int j=0;j<M;j++)cin>>field[i][j];
    cin>>P>>Q;
    for(int i=0;i<P;i++)for(int j=0;j<Q;j++)cin>>pattern[i][j];
    compute_hash(pattern,P,Q);
    ull ttmp=hash_[0][0];
    compute_hash(field,N,M);
    for(int i=0;i+P<=N;i++){
        for(int j=0;j+Q<=M;j++){
            if(hash_[i][j]==ttmp){
                cout<<i<<" "<<j<<endl;
            }
        }
    }
    //-------------------------------  
    //ll end_time=clock();cout<<"time="<<end_time-begin_time<<"ms"<<endl;
    //-------------------------------
    return 0;
}
//----------------------------------------------------------------------














































### Prompt

Please create a solution in Cpp to the following problem:

Find places where a R × C pattern is found within a H × W region. Print top-left coordinates (i, j) of sub-regions where the pattern found. The top-left and bottom-right coordinates of the region is (0, 0) and (H-1, W-1) respectively.

Constraints

* 1 ≤ H, W ≤ 1000
* 1 ≤ R, C ≤ 1000
* The input consists of alphabetical characters and digits

Input

In the first line, two integers H and W are given. In the following H lines, i-th lines of the region are given.

In the next line, two integers R and C are given. In the following R lines, i-th lines of the pattern are given.

Example

Input

4 5
00010
00101
00010
00100
3 2
10
01
10


Output

0 3
1 2

### Response

```cpp
#pragma GCC optimize ("O3")
#include <iostream>
#include <iomanip>
#include <istream>
#include <ostream>
#include <sstream>
#include <iterator>
#include <vector>
#include <algorithm>
#include <queue>
#include <deque>
#include <list>
#include <stack>
#include <map>
#include <unordered_map>
#include <set>
#include <bitset>
#include <utility>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <string>
#include <ctime>
#include <cctype>
#include <cstdlib>
#define IINF 10e8
#define INF 1<<30
#define MOD 1000000007
#define mod 1000000007
#define REP(i, a, n) for (ll i = a; i < (ll)(n); i++)
#define REPE(i, a, n) for (ll i = a; i <= (ll)(n); i++)
#define Endl endl
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define mt make_tuple
#define eb emplace_back
#define mmax(x,y)(x>y?x:y)
#define mmin(x,y)(x<y?x:y)
#define chmax(x,y) x=mmax(x,y)
#define chmin(x,y) x=mmin(x,y)
#define all(x) (x).begin(),(x).end()
#define siz(x) (ll)(x).size()
#define PI acos(-1.0)
using namespace std;
typedef long long int ll;
typedef unsigned long long int ull;
typedef long double ld;
typedef pair<int,int>Pin;
typedef pair<ll,ll>Pll;
template<class T> using V=vector<T>;
long long GCD(long long a, long long b) {return b?GCD(b,a%b):a;}
long long LCM(long long a, long long b) {return a/GCD(a,b)*b;}
int dx[4]={-1,0,1,0};
int dy[4]={0,-1,0,1};
int ddx[8]={-1,0,1,0,1,1,-1,-1};
int ddy[8]={0,-1,0,1,1,-1,1,-1};
ll cmp(pair<ll,ll>a,pair<ll,ll> b){
        if(a.se!=b.se)
        return a.se<b.se;
        else
        return a.fi<b.fi;
}
//----------------------------------------------------------------------
int N,M,T,P,Q;
char field[1010][1010];
char pattern[1010][1010];
ll hash_[1010][1010],tmp[1010][1010];

void compute_hash(char a[1010][1010],int n,int m){
    const ull B1=9973;
    const ull B2=100000007;
    ull t1=1;
    for(int j=0;j<Q;j++)t1*=B1;
    for(int i=0;i<n;i++){
        ull e=0;
        for(int j=0;j<Q;j++)e=e*B1+a[i][j];
        for(int j=0;j+Q<=m;j++){
            tmp[i][j]=e;
            if(j+Q<m)e=e*B1-t1*a[i][j]+a[i][j+Q];
        }
    }
    ull t2=1;
    for(int i=0;i<P;i++)t2*=B2;
    for(int j=0;j+Q<=m;j++){
        ull e=0;
        for(int i=0;i<P;i++)e=e*B2+tmp[i][j];
        for(int i=0;i+P<=n;i++){
            hash_[i][j]=e;
            if(i+P<n)e=e*B2-t2*tmp[i][j]+tmp[i+P][j];
        }
    }
}
//----------------------------------------------------------------------
int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    //-------------------------------  
    //ll begin_time=clock();
    //-------------------------------
    cin>>N>>M;
    for(int i=0;i<N;i++)for(int j=0;j<M;j++)cin>>field[i][j];
    cin>>P>>Q;
    for(int i=0;i<P;i++)for(int j=0;j<Q;j++)cin>>pattern[i][j];
    compute_hash(pattern,P,Q);
    ull ttmp=hash_[0][0];
    compute_hash(field,N,M);
    for(int i=0;i+P<=N;i++){
        for(int j=0;j+Q<=M;j++){
            if(hash_[i][j]==ttmp){
                cout<<i<<" "<<j<<endl;
            }
        }
    }
    //-------------------------------  
    //ll end_time=clock();cout<<"time="<<end_time-begin_time<<"ms"<<endl;
    //-------------------------------
    return 0;
}
//----------------------------------------------------------------------














































```