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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.Arrays;
import java.util.HashSet;
import java.util.Scanner;
import java.util.Stack;

public class Main 
{
	static class Node
	{
		int i;
		HashSet<Node> e = new HashSet<>();
		boolean vis = false;
		
		public Node(int id)
		{
			i=id;
		}
		
		public String toString()
		{
			return Integer.toString(i);
		}
	}

	static Node[] nd;
	
	public static void main(String[] args) 
	{
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int m = sc.nextInt();
		nd = new Node[n];
		boolean[] rootcan = new boolean[n];
		
		for(int i = 0; i < n; i++)
		{
			nd[i] = new Node(i);
		}

		for(int i = 0; i < n-1+m; i++)
		{
			int a = sc.nextInt()-1;
			int b = sc.nextInt()-1;
			nd[a].e.add(nd[b]);
			rootcan[b] = true;
		}
		
		int root = -1;
		for(int i = 0; i < n && root==-1; i++)
			if(!rootcan[i])
				root = i;
		
		Stack<Node> q = new Stack<>();
		Stack<Node> st = new Stack<>();
		
		q.add(nd[root]);
		
		while(!q.isEmpty())
		{
			Node pos = q.peek();
			
			if(!pos.vis)
			{
				pos.vis = true;
				
				for(Node e : pos.e)
				{
					if(!e.vis)
					{
						q.add(e);
					}
				}
			}
			else
			{ //backtrack
				st.add(pos);
//				System.out.println("bt " + pos);
				q.pop();
			}
		}
		
//		System.out.println(st);
		int[] a = new int[n];
		for(int i = 0; i < n-1; i++)
		{
//			System.out.println("peek " + st.peek());
//			System.out.println("peek " + Integer.toString((int)(st.peek().i)+1));
			Node pos = st.pop();
			for(Node e : pos.e)
			{
				a[e.i] = pos.i+1;
			}
		}
		
		for(int i = 0; i < n; i++)
			System.out.println(a[i]);
//		System.out.println(Arrays.toString(a));

	}

}