Problem:
Score : 500 points

Problem Statement

There is a rooted tree (see Notes) with N vertices numbered 1 to N.
Each of the vertices, except the root, has a directed edge coming from its parent.
Note that the root may not be Vertex 1.

Takahashi has added M new directed edges to this graph.
Each of these M edges, u \rightarrow v, extends from some vertex u to its descendant v.

You are given the directed graph with N vertices and N-1+M edges after Takahashi added edges.
More specifically, you are given N-1+M pairs of integers, (A_1, B_1), ..., (A_{N-1+M}, B_{N-1+M}), which represent that the i-th edge extends from Vertex A_i to Vertex B_i.

Restore the original rooted tree.

Notes

For "tree" and other related terms in graph theory, see the article in Wikipedia, for example.

Constraints

3 \leq N

1 \leq M

N + M \leq 10^5

1 \leq A_i, B_i \leq N

A_i \neq B_i

If i \neq j, (A_i, B_i) \neq (A_j, B_j).

The graph in input can be obtained by adding M edges satisfying the condition in the problem statement to a rooted tree with N vertices.

Input

Input is given from Standard Input in the following format:

N M
A_1 B_1
:
A_{N-1+M} B_{N-1+M}

Output

Print N lines.
In the i-th line, print 0 if Vertex i is the root of the original tree, and otherwise print the integer representing the parent of Vertex i in the original tree.

Note that it can be shown that the original tree is uniquely determined.

Sample Input 1

3 1
1 2
1 3
2 3

Sample Output 1

0
1
2

The graph in this input is shown below:

It can be seen that this graph is obtained by adding the edge 1 \rightarrow 3 to the rooted tree 1 \rightarrow 2 \rightarrow 3.

Sample Input 2

6 3
2 1
2 3
4 1
4 2
6 1
2 6
4 6
6 5

Sample Output 2

6
4
2
0
6
2

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <queue>
#include <vector>

using namespace std;

void solve() {
  int N, M;
  cin >> N >> M;

  vector<vector<int>> V(N+1);
  vector<int> income(N+1);
  vector<int> parent(N+1);


  for (int i = 0; i < N-1+M; i++) {
    int a, b; cin >> a >> b;
    V[a].push_back(b);
    income[b]++;
  }

  int root = 1;
  while (income[root] > 0) {
    root++;
  }

  parent[root] = 0;

  queue<int> q;
  q.push(root);

  while (!q.empty()) {
    int x = q.front(); q.pop();

    for (int dist : V[x]) {
      income[dist]--;
      if (income[dist] == 0) {
        q.push(dist);
        parent[dist] = x;
      }
    }
  }

  for (int i = 1; i <= N; i++) {
    cout << parent[i] << endl;
  }
}

signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  solve();
  return 0;
}