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.io.*; 
import java.util.*; 
  
class Graph { 
    ArrayList<Integer> adjArr[];
  
    @SuppressWarnings("unchecked")
	Graph(int v) {
        adjArr = new ArrayList[v]; 
        for (int i=0; i < v; ++i) 
            adjArr[i] = new ArrayList<Integer>(); 
    } 
  
    // Add directed pair to graph
    void addToGraph(int id, int neighbor) { 
    	adjArr[id].add(neighbor); 
    	}
    
    // Topologically sorts adjacency list graph into a stack
    void topsort(int v, boolean visited[], Stack<Integer> stack) {  
        // Mark current as visited
    	visited[v] = true; 
        int i; 

        Iterator<Integer> it = adjArr[v].iterator(); 
        while (it.hasNext()) { 
            i = it.next(); 
            // Topsort through all unvisited adjacent vertices before adding to stack
            if (!visited[i]) 
                topsort(i, visited, stack); 
        } 
        stack.push(v); 
    }
}

public class Main {
  
    public static void main(String args[]) 
    { 
    	Scanner in = new  Scanner(System.in);

		int totalVertex = in.nextInt();
		Graph adj = new Graph(totalVertex);
		
		int totalConnections = in.nextInt();
		
		int id, neighbor;
		
		for (int i = 0; i < totalConnections; i++) {
			id = in.nextInt();
			neighbor = in.nextInt();
			adj.addToGraph(id, neighbor);
		}
		
		Stack<Integer> stack = new Stack<Integer>();
		
		boolean[] visited = new boolean[totalVertex];
		// Initialize visited array with false
		for(int i = 0; i < totalVertex; i++){
			visited[i] = false;
		}
		
		for(int i = 0; i < totalVertex; i++) {
			if(!visited[i]) {
				adj.topsort(i, visited, stack);
			}
		}
		
		while(!stack.isEmpty()) {
			System.out.println(stack.pop());
		}
    }
}