Problem: <image>


A directed acyclic graph (DAG) can be used to represent the ordering of tasks. Tasks are represented by vertices and constraints where one task can begin before another, are represented by edges. For example, in the above example, you can undertake task B after both task A and task B are finished. You can obtain the proper sequence of all the tasks by a topological sort.

Given a DAG $G$, print the order of vertices after the topological sort.

Constraints

* $1 \leq |V| \leq 10,000$
* $0 \leq |E| \leq 100,000$
* There are no parallel edges in $G$
* There are no self loops in $G$

Input

A directed graph $G$ is given in the following format:
$|V|\;|E|$
$s_0 \; t_0$
$s_1 \; t_1$
:
$s_{|E|-1} \; t_{|E|-1}$



$|V|$ is the number of vertices and $|E|$ is the number of edges in the graph. The graph vertices are named with the numbers $0, 1,..., |V|-1$ respectively.

$s_i$ and $t_i$ represent source and target nodes of $i$-th edge (directed).

Output

Print the vertices numbers in order. Print a number in a line.

If there are multiple possible solutions, print any one of them (the solution is judged by a special validator).

Example

Input

6 6
0 1
1 2
3 1
3 4
4 5
5 2


Output

0
3
1
4
5
2

Solution (in JAVA):

Solution:
import java.util.LinkedList;
import java.util.Scanner;
import java.util.Stack;

class Node {
	public int id;
	public LinkedList<Node> sources;
	public LinkedList<Node> targets;
	
	public Node(int id) {
		this.id = id;
		sources = new LinkedList<>();
		targets = new LinkedList<>();
	}
}

public class Main {
	public static boolean[] visited = null;
	
	public static Stack<Integer> topologicalSort(Stack<Node> nodes, int overallLength) {
		Stack<Integer> result = new Stack<Integer>();
		boolean[] visited = new boolean[overallLength];
		
		//go up the graph
		while (!nodes.isEmpty()) {
			Node currentNode = nodes.pop();
			result.push(currentNode.id);
			visited[currentNode.id] = true;
			
			//go through sources. If we have visited all the targets, we can add it to stack
			for (int i = 0; i < currentNode.sources.size(); i++) {
				Node currentSource = currentNode.sources.get(i);
				boolean isGood = true;
				for (int j = 0; j < currentSource.targets.size(); j++) {
					if (!visited[currentSource.targets.get(j).id]) {
						isGood = false;
						break;
					}
				}
				
				if (isGood) nodes.push(currentSource);
			}
		}
		
		return result;
	}
	
	public static void main(String[] args) {
		Scanner scanner = new Scanner(System.in);
		int numVerticies = scanner.nextInt();
		int numEdges = scanner.nextInt();
		Node[] nodes = new Node[numVerticies];
		visited = new boolean[numVerticies];
		
		//creating all nodes as empty
		for (int i = 0; i < numVerticies; i++) {
			nodes[i] = new Node(i);
		}
		
		//creating all relations to nodes
		for (int i = 0; i < numEdges; i++) {
			int source = scanner.nextInt();
			int target = scanner.nextInt();
			
			nodes[source].targets.add(nodes[target]);
			nodes[target].sources.add(nodes[source]);
		}
		scanner.close();
		
		//create Stack with only nodes that doesn't point to anything
		Stack<Node> rootNodes = new Stack<>();
		for (int i = 0; i < nodes.length; i++) {
			if (nodes[i].targets.isEmpty()) {
				rootNodes.push(nodes[i]);
			}
		}
		
		//topological sort
		Stack<Integer> results = topologicalSort(rootNodes, nodes.length);
		while (!results.isEmpty()) {
			System.out.println(results.pop());
		}
	}
}