#include <iostream>
#include <algorithm>
#include <vector>

using namespace std;

int main() {
	int n, m; cin >> n >> m;
	vector<vector<int> > G(n, vector<int>(n, 1));
	for (int i = 0; i < n; ++i) G[i][i] = 0;
	while (m--) {
		int a, b; cin >> a >> b;
		G[a - 1][b - 1] = 0;
	}
	for (int k = 0; k < n; ++k)
		for (int i = 0; i < n; ++i)
			for (int j = 0; j < n; ++j)
				G[i][j] = min(G[i][j], G[i][k] + G[k][j]);
	for (int i = 0; i < n; ++i) {
		bool flag = false;
		for (int j = 0; j < n; ++j) {
			if (G[i][j] + G[j][i] == 0) {
				if (flag) cout << " ";
				cout << j + 1, flag = true;
			}
		}
		cout << endl;
	}

	return 0;
}

### Prompt

Create a solution in Cpp for 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 <iostream>
#include <algorithm>
#include <vector>

using namespace std;

int main() {
	int n, m; cin >> n >> m;
	vector<vector<int> > G(n, vector<int>(n, 1));
	for (int i = 0; i < n; ++i) G[i][i] = 0;
	while (m--) {
		int a, b; cin >> a >> b;
		G[a - 1][b - 1] = 0;
	}
	for (int k = 0; k < n; ++k)
		for (int i = 0; i < n; ++i)
			for (int j = 0; j < n; ++j)
				G[i][j] = min(G[i][j], G[i][k] + G[k][j]);
	for (int i = 0; i < n; ++i) {
		bool flag = false;
		for (int j = 0; j < n; ++j) {
			if (G[i][j] + G[j][i] == 0) {
				if (flag) cout << " ";
				cout << j + 1, flag = true;
			}
		}
		cout << endl;
	}

	return 0;
}

```