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: Time Limit Exceeded)

#include "iostream"
#include "vector"
#include "string"
#include "algorithm"
#include "math.h"
#include "cstring"
#include "queue"

using namespace std;
typedef long long ll;

int main()
{
  int n, m;
  cin >> n >> m;
  int a[100100], b[100100];
  vector<int> neigh[100100];
  for (int i = 0; i < n + m - 1; i++)
  {
    cin >> a[i] >> b[i];
    neigh[a[i]].push_back(b[i]);
  }

  vector<int> far[100100];
  for (int i = 1; i <= n; i++)
  {
    for (int j = 0; j < neigh[b[i]].size(); j++)
    {
      far[a[i]].push_back(neigh[b[i]][j]);
    }
  }

  for (int loop = 0; loop < 100100; loop++)
  {
    for (int i = 1; i <= n; i++)
    {
      for (int j = 0; j < far[b[i]].size(); j++)
      {
        auto result = find(far[a[i]].begin(), far[a[i]].end(), far[b[i]][j]);
        if (result == far[a[i]].end())
        {
          far[a[i]].push_back(far[b[i]][j]);
        }
      }
    }
  }

  int ans[100100]={0};

  for (int i = 1; i <= n; i++)
  {
    for (int j = 1; j <= n; j++)
    {
      auto result1 = find(neigh[i].begin(), neigh[i].end(), j);
      auto result2 = find(far[i].begin(), far[i].end(), j);
      if (result1 != neigh[i].end() && result2 == far[i].end())
      {
        ans[j]=i;
      }
    }
  }

  for (int i = 1; i <= n; i++)
  {
   cout<< ans[i] << "\n";
  }
}