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.ArrayList;
import java.util.Arrays;
import java.util.LinkedList;
import java.util.Scanner;

public class Main {
	MyScanner sc = new MyScanner();
	Scanner sc2 = new Scanner(System.in);
	long start = System.currentTimeMillis();
	long fin = System.currentTimeMillis();
	final int MOD = 1000000007;
	int[] dx = { 1, 0, 0, -1 };
	int[] dy = { 0, 1, -1, 0 };

	int N, M;
	int[] indeg;
	boolean[] V;
	ArrayList<Integer>[] G;
	ArrayList<Integer> res;

	@SuppressWarnings("unchecked")
	void run() {
		N = sc.nextInt();
		M = sc.nextInt();
		indeg = new int[N];
		G = new ArrayList[N];
		V = new boolean[N];
		res = new ArrayList<Integer>();
		for (int i = 0; i < N; i++)
			G[i] = new ArrayList<Integer>();
		for (int i = 0; i < M; i++) {
			int s = sc.nextInt();
			int t = sc.nextInt();
			G[s].add(t);
			indeg[t]++;
		}
		topologicalSort();
	}

	void topologicalSort() {
		for (int u = 0; u < N; u++) {
			if (indeg[u] == 0 && !V[u]) {
				bfs(u);
			}
		}
		StringBuilder out = new StringBuilder();
		for (int i = 0; i < N; i++) out.append(res.get(i) + "\n");
		System.out.print(out);
	}

	void bfs(int s) {
		LinkedList<Integer> que = new LinkedList<Integer>();
		que.add(s);
		V[s] = true;
		while (!que.isEmpty()) {
			int u = que.poll();
			res.add(u);
			for (int i = 0; i < G[u].size(); i++) {
				int v = G[u].get(i);
				indeg[v]--;
				if (indeg[v] == 0 && !V[v]) {
					que.add(v);
					V[v] = true;
				}
			}
		}
	}

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

	void debug(Object... o) {
		System.out.println(Arrays.deepToString(o));
	}

	void debug2(char[][] array) {
		for (int i = 0; i < array.length; i++) {
			for (int j = 0; j < array[i].length; j++) {
				System.out.print(array[i][j]);
			}
			System.out.println();
		}
	}

	boolean inner(int h, int w, int limH, int limW) {
		return 0 <= h && h < limH && 0 <= w && w < limW;
	}

	void swap(char[] x, int a, int b) {
		char tmp = x[a];
		x[a] = x[b];
		x[b] = tmp;
	}

	// find minimum i (k <= a[i])
	int lower_bound(int a[], int k) {
		int l = -1;
		int r = a.length;
		while (r - l > 1) {
			int mid = (l + r) / 2;
			if (k <= a[mid])
				r = mid;
			else
				l = mid;
		}
		// r = l + 1
		return r;
	}

	// find minimum i (k < a[i])
	int upper_bound(int a[], int k) {
		int l = -1;
		int r = a.length;
		while (r - l > 1) {
			int mid = (l + r) / 2;
			if (k < a[mid])
				r = mid;
			else
				l = mid;
		}
		// r = l + 1
		return r;
	}

	long gcd(long a, long b) {
		return a % b == 0 ? b : gcd(b, a % b);
	}

	long lcm(long a, long b) {
		return a * b / gcd(a, b);
	}

	boolean palindrome(String s) {
		for (int i = 0; i < s.length() / 2; i++) {
			if (s.charAt(i) != s.charAt(s.length() - 1 - i)) {
				return false;
			}
		}
		return true;
	}

	class MyScanner {
		int nextInt() {
			try {
				int c = System.in.read();
				while (c != '-' && (c < '0' || '9' < c))
					c = System.in.read();
				if (c == '-')
					return -nextInt();
				int res = 0;
				do {
					res *= 10;
					res += c - '0';
					c = System.in.read();
				} while ('0' <= c && c <= '9');
				return res;
			} catch (Exception e) {
				return -1;
			}
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		String next() {
			try {
				StringBuilder res = new StringBuilder("");
				int c = System.in.read();
				while (Character.isWhitespace(c))
					c = System.in.read();
				do {
					res.append((char) c);
				} while (!Character.isWhitespace(c = System.in.read()));
				return res.toString();
			} catch (Exception e) {
				return null;
			}
		}

		int[] nextIntArray(int n) {
			int[] in = new int[n];
			for (int i = 0; i < n; i++) {
				in[i] = nextInt();
			}
			return in;
		}

		int[][] nextInt2dArray(int n, int m) {
			int[][] in = new int[n][m];
			for (int i = 0; i < n; i++) {
				in[i] = nextIntArray(m);
			}
			return in;
		}

		double[] nextDoubleArray(int n) {
			double[] in = new double[n];
			for (int i = 0; i < n; i++) {
				in[i] = nextDouble();
			}
			return in;
		}

		long[] nextLongArray(int n) {
			long[] in = new long[n];
			for (int i = 0; i < n; i++) {
				in[i] = nextLong();
			}
			return in;
		}

		char[][] nextCharField(int n, int m) {
			char[][] in = new char[n][m];
			for (int i = 0; i < n; i++) {
				String s = sc.next();
				for (int j = 0; j < m; j++) {
					in[i][j] = s.charAt(j);
				}
			}
			return in;
		}
	}
}