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.io.PrintWriter;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.List;
import java.util.NoSuchElementException;
import java.util.Vector;

public class Main {

	public static void main(String[] args) throws IOException {
		new Main().solve();
	}

	private void solve() throws IOException {
		try {
			solveA();
		} finally {
			if (in != null) {
				in.close();
			}
			if (out != null) {
				out.flush();
				out.close();
			}
		}

	}

	private void solveA() {
		int v = nextInt();
		int e = nextInt();
		//		List<Integer>[] adj = new List[v];
		List<List<Integer>> adj = new ArrayList<List<Integer>>();
		for (int i = 0; i < v; i++) {
			adj.add(new ArrayList<Integer>());
		}
		for (int i = 0; i < e; i++) {
			int from = nextInt();
			int to = nextInt();
			//			adj[from].add(to);
			adj.get(from).add(to);
		}
		// Create a array to store indegrees of all
		// vertices. Initialize all indegrees as 0.
		int indegree[] = new int[v];

		// Traverse adjacency lists to fill indegrees of
		// vertices. This step takes O(V+E) time
		for (int i = 0; i < v; i++) {
			//			List<Integer> temp = adj[i];
			List<Integer> temp = adj.get(i);
			for (int node : temp) {
				indegree[node]++;
			}
		}

		// Create a queue and enqueue all vertices with
		// indegree 0
		ArrayDeque<Integer> q = new ArrayDeque<Integer>();
		for (int i = 0; i < v; i++) {
			if (indegree[i] == 0)
				q.addLast(i);
		}

		// Initialize count of visited vertices
		int cnt = 0;

		// Create a vector to store result (A topological
		// ordering of the vertices)
		Vector<Integer> topOrder = new Vector<Integer>();
		while (!q.isEmpty()) {
			// Extract front of queue (or perform dequeue)
			// and add it to topological order
			int u = q.removeFirst();
			topOrder.add(u);

			// Iterate through all its neighbouring nodes
			// of dequeued node u and decrease their in-degree
			// by 1
			//			for (int node : adj[u]) {
			for (int node : adj.get(u)) {
				// If in-degree becomes zero, add it to queue
				if (--indegree[node] == 0)
					q.addFirst(node);
			}
			cnt++;
		}

		// Check if there was a cycle
		if (cnt != v) {
			System.out.println("There exists a cycle in the graph");
			return;
		}

		// Print topological order
		for (int i : topOrder) {
			out.println(i);
		}
	}

	private final PrintWriter out = new PrintWriter(System.out);
	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 int nextInt() {
		return Integer.parseInt(next());
	}

	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() {
		if (!hasNext())
			throw new NoSuchElementException();
		long n = 0;
		boolean minus = false;
		int b = readByte();
		if (b == '-') {
			minus = true;
			b = readByte();
		}
		if (b < '0' || '9' < b) {
			throw new NumberFormatException();
		}
		while (true) {
			if ('0' <= b && b <= '9') {
				n *= 10;
				n += b - '0';
			} else if (b == -1 || !isPrintableChar(b)) {
				return minus ? -n : n;
			} else {
				throw new NumberFormatException();
			}
			b = readByte();
		}
	}
}