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;

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<List<Integer>> rinsetuList = new ArrayList<List<Integer>>();

		for (int i = 0; i < v; i++) {
			rinsetuList.add(new ArrayList<Integer>());
		}

		for (int i = 0; i < e; i++) {
			rinsetuList.get(nextInt()).add(nextInt());
		}

		int[] indegrees = new int[v];
		for (List<Integer> node : rinsetuList) {
			for (Integer i : node) {
				indegrees[i]++;
			}
		}

		ArrayDeque<Integer> queue = new ArrayDeque<Integer>();
		for (int i = 0; i < indegrees.length; i++) {
			if (indegrees[i] == 0) {
				queue.addLast(i);
			}
		}

		List<Integer> res = new ArrayList<Integer>();
		int cnt = 0;
		while (queue.size() != 0) {
			int index = queue.removeLast();
			res.add(index);

			for (Integer nextDegree : rinsetuList.get(index)) {
				int nextCnt = --indegrees[nextDegree];
				if (nextCnt == 0) {
					queue.addLast(nextDegree);
				}
			}
			if (cnt > v) {
				out.println("This graph includes the Cycle");
			}
		}

		for (Integer integer : res) {
			out.println(integer);
		}

	}

	/**
	 * http://www.thothchildren.com/chapter/5bcc8bc051d9305189030f9f
	 * https://tubo28.me/algorithm/tsort/
	 * http://www.dais.is.tohoku.ac.jp/~shioura/teaching/ad09/ad09-13.pdf
	 * https://www.geeksforgeeks.org/topological-sorting-indegree-based-solution/
	 * トポロジカルソート
	 */
	private void solveA2() {
		int v = nextInt();
		int e = nextInt();

		/*
		 *隣接リストの作成
		 */
		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.get(from).add(to);
		}
		// 入次数が0のものを判定するための配列
		int indegree[] = new int[v];
		// 入次数0を判定
		for (int i = 0; i < v; i++) {
			List<Integer> temp = adj.get(i);
			//iをfromとするnode達
			for (int node : temp) {
				//入次数の個数
				indegree[node]++;
			}
		}

		/*
		 * queueの作成
		 * 入次数0のものをqueueに詰める
		 * 入次数0から調査していく
		 */
		ArrayDeque<Integer> q = new ArrayDeque<Integer>();
		for (int i = 0; i < v; i++) {
			if (indegree[i] == 0) {
				q.addLast(i);
			}
		}

		//訪問済み頂点数
		int cnt = 0;

		// トポロジカルソートの結果
		List<Integer> res = new ArrayList<Integer>();

		/*
		 * BFS
		 */
		while (!q.isEmpty()) {
			// 接続先の頂点を探索開始
			int u = q.removeFirst();
			//入次数0なのでリザルトにadd
			res.add(u);

			/*
			 * この頂点の次の接続先の入次数を-する
			 * その結果、入次数=0となる場合はソートリザルトに追加し、次の探索に利用する
			 */
			for (int node : adj.get(u)) {

				indegree[node]--;
				if (indegree[node] == 0) {
					q.addFirst(node);
				}
			}
			cnt++;
			if (cnt > v) {
				System.out.println("graph内に循環有");
				return;
			}
		}

		for (int i : res) {
			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();
		}
	}
}