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.IOException;
import java.io.InputStream;
import java.util.ArrayList;
import java.util.HashSet;
import java.util.NoSuchElementException;
 
public class Main {
	private static FastScanner sc = new FastScanner();
	private static Node[] V;
	private static ArrayList<Integer> sort = new ArrayList<>();
	
	public static void main(String[] args) {
		int Vn = sc.nextInt();
		int En = sc.nextInt();
		
		V = new Node[Vn];
		for(int i=0; i<Vn; i++) {
			V[i] = new Node();
			V[i].id = i;
		}
		
		for(int i=0; i<En; i++) {
			int s = sc.nextInt();
			int t = sc.nextInt();
			V[s].nbr.add(t);
		}
		
		for(int i=0; i<Vn; i++) {
			if(V[i].visited == 0) {
				V[i].visit();
			}
		}
		
		for(int i=sort.size()-1; i>=0; i--) {
			System.out.println(sort.get(i));
		}
	}
	
	static class Node {
		int id;
		HashSet<Integer> nbr = new HashSet<>();
		int visited = 0;
		
		void visit() {
			if(visited == 1) {
				System.out.println(1);
				System.exit(0);
			} else if(visited == 0) {
				visited = 1;
				for(int x : nbr) {
					V[x].visit();
				}
				
				visited = 2;
				sort.add(id);
			} 
		}
	}
	
	static class FastScanner {
        private final InputStream in = System.in;
        private final byte[] buffer = new byte[1024];
        private int ptr = 0;
        private int buflen = 0;
        private boolean hasNextByte() {
            if(ptr < buflen) {
                return true;
            } else {
                ptr = 0;
                try {
                    buflen = in.read(buffer);
                } catch(IOException e) {
                    e.printStackTrace();
                }
                if(buflen <= 0) {
                    return false;
                }
            }
            return true;
        }
        private int readByte() { if (hasNextByte()) return buffer[ptr++]; else return -1;}
        private static boolean isPrintableChar(int c) { return 33 <= c && c <= 126;}
        private void skipUnprintable() { while(hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++;}
        public boolean hasNext() { skipUnprintable(); return hasNextByte();}
        public String next() {
            if (!hasNext()) throw new NoSuchElementException();
            StringBuilder sb = new StringBuilder();
            int b = readByte();
            while(isPrintableChar(b)) {
                sb.appendCodePoint(b);
                b = readByte();
            }
            return sb.toString();
        }
        public long nextLong() {
            return Long.parseLong(next());
        }
        public int nextInt(){
            return Integer.parseInt(next());
        }
        public double nextDouble(){
            return Double.parseDouble(next());
        }
    }
}