#include <bits/stdc++.h>
using namespace std;

int main(void){
    int n, m;
    cin >> n >> m;
    vector<vector<int> > d(n, vector<int>(n,1));
    for(int i=0;i<n;i++){
        d[i][i] = 0;
    }
    for(int i=0;i<m;i++){
        int a, b;
        cin >> a >> b;
        a--; b--;
        d[a][b] = 0;
    }
    for(int k=0;k<n;k++){
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                d[i][j] = min(d[i][j], d[i][k]+d[k][j]);
            }
        }
    }
    for(int i=0;i<n;i++){
        bool f = false;
        for(int j=0;j<n;j++){
            if(d[i][j] || d[j][i]) continue;
            if(f) cout << " ";
            f = true;
            cout << j+1;
        }
        cout << endl;
    }
    return 0;
}

### Prompt

Please formulate a Cpp solution to the following problem:

problem

There are $ N $ propositions, named $ 1, 2, \ cdots, N $, respectively. Also, $ M $ information about the propositions is given. The $ i $ th information is "$ a_i $$". Given in the form "b_i $", which means that $ a_i $ is $ b_i $. ("If" is a logical conditional and the transition law holds.) $ For each proposition $ i $ Output all propositions that have the same value as i $ in ascending order. However, proposition $ i $ and proposition $ i $ are always the same value. Proposition $ X $ and proposition $ Y $ have the same value as "$ if $ X $". It means "Y $" and "$ X $ if $ Y $".



output

On the $ i $ line, output all propositions that have the same value as the proposition $ i $, separated by blanks in ascending order. Also, output a line break at the end of each line.

Example

Input

5 2
1 2
2 1


Output

1 2
1 2
3
4
5

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;

int main(void){
    int n, m;
    cin >> n >> m;
    vector<vector<int> > d(n, vector<int>(n,1));
    for(int i=0;i<n;i++){
        d[i][i] = 0;
    }
    for(int i=0;i<m;i++){
        int a, b;
        cin >> a >> b;
        a--; b--;
        d[a][b] = 0;
    }
    for(int k=0;k<n;k++){
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                d[i][j] = min(d[i][j], d[i][k]+d[k][j]);
            }
        }
    }
    for(int i=0;i<n;i++){
        bool f = false;
        for(int j=0;j<n;j++){
            if(d[i][j] || d[j][i]) continue;
            if(f) cout << " ";
            f = true;
            cout << j+1;
        }
        cout << endl;
    }
    return 0;
}

```