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: Time Limit Exceeded)

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

public class Main {
	static int N;
	static int M;
	static Node[] nodeArray;
	static Node root;

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

		solve();
	}

	static void solve() {
		boolean[] isSearched = new boolean[N+1];
		Deque<Integer> sortedNode = new ArrayDeque<>(N+1);
		
		for(int i = 1; i<=N; i++) {
			recursiveSearch(i, isSearched, sortedNode);
		}
		
		Integer[] sortedNodeArray = new Integer[sortedNode.size()];
		sortedNode.toArray(sortedNodeArray);
		
		int[] parentList = new int[N+1];
		
		for(int i = sortedNodeArray.length - 1; i>=0; i--) {
			int nodeId = sortedNodeArray[i];
			
			for(int j = i-1; j>=0; j--) {
				if(nodeArray[nodeId].parents.contains(sortedNodeArray[j])) {
					parentList[nodeId] = sortedNodeArray[j];
					break;
				}
			}
		}
		
		for(int i = 1; i<=N; i++) {
			System.out.println(parentList[i]);
		}
	}
	
	static void recursiveSearch(int nodeId, boolean[] isSearched, Deque<Integer> queue) {
		if(isSearched[nodeId]) {
			return;
		}
		
		isSearched[nodeId] = true;
		
		for(int childId : nodeArray[nodeId].children) {
			recursiveSearch(childId, isSearched, queue);
		}
		
		queue.addFirst(nodeId);
	}

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

			nodeArray = new Node[N+1];

			for(int i = 0; i<nodeArray.length; i++) {
				nodeArray[i] = new Node(i);
			}

			// 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[d].parents.add(s);
				nodeArray[s].children.add(d);
			}
		}
	}

	static class Node{
		public int id;
		public ArrayList<Integer> parents;
		public ArrayList<Integer> children;

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