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.ArrayDeque;
import java.util.ArrayList;
import java.util.Queue;
import java.util.Scanner;

public class Main {
	public static final int MAX_HIGHT = 100;

	static int N;
	static int M;
	static Node[] nodeArray;
	static Node root;

	public static void main(String[] args) {
		readInput();

		solve();
	}

	public static void solve() {
		// 幅優先探索で距離が長いパスがもともとのパス
		Queue<Node> queue = new ArrayDeque<>();
		queue.add(root);
		
		boolean[] isSearched = new boolean[N+1];

		while(!queue.isEmpty()) {
			Node node = queue.poll();
			
			// 探索済みなら探索不要
			if(isSearched[node.id]) {
				continue;
			}
			
			for(Node child : node.children) {
				child.parentId = node.id;
				queue.add(child);
			}
			isSearched[node.id] = true;
		}
		
		for(int i = 1; i<=N; i++) {
			System.out.println(nodeArray[i].parentId);
		}
	}

	public static void readInput() {
		try (Scanner sc = new Scanner(System.in)) {
			N = sc.nextInt();
			M = sc.nextInt();

			nodeArray = new Node[N+1];

			for(int i = 1; i<nodeArray.length; i++) {
				nodeArray[i] = new Node(i);
			}
			
			// 根（親のいないノード）の検索用
			boolean[] hasParent = new boolean[N+1];
			
			// 辺の情報を読み込む
			for(int i = 0; i<(N - 1 + M); i++) {
				int s = sc.nextInt();
				int d = sc.nextInt();

				nodeArray[s].children.add(nodeArray[d]);
				hasParent[d] = true;
			}
			
			// 根を探す
			for(int i = 1; i<=N; i++) {
				if(!hasParent[i]) {
					root = nodeArray[i];
				}
			}
		}
	}
	
	public static class Node{
		public int id;
		public int parentId;
		public ArrayList<Node> children;

		public Node(int id) {
			parentId = 0;
			this.id = id;
			children = new ArrayList<Node>();
		}
	}
}