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: Wrong Answer)

#include<bits/stdc++.h>

using namespace std;

const int N = 100005;

vector<int> g[N];
int n, m, vis[N], par[N], in[N];

void dfs(int u, int p) {
  par[u] = p;
  vis[u] = 1;
  for (auto v : g[u]) {
    if (vis[v]) {
      if (par[v] == 0) par[v] = u;
    } else {
      dfs(v, u);
    }
  }
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 0; i < m + (n - 1); i++) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    in[v]++;
  }
  int mn = 1e9, start = 1;
  for (int i = 1; i <= n; i++) {
    if (mn > in[i]) {
      start = i;
      mn = in[i];
    }
  }
  //cout << "S: " << start << endl;
  dfs(start, 0);
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) {
      dfs(i, 0);
    }
  }
  for (int i = 1; i <= n; i++) {
    cout << par[i] << '\n';
  }
  return 0;
}