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.*;

public class Main {

    private static int MOD = 1_000_000_007;

    public static void main(String[] args) {
        final Scanner scanner = new Scanner(System.in);
        solve(scanner);
    }

    static void solve(Scanner scanner) {
        int V = scanner.nextInt();
        int E = scanner.nextInt();
        scanner.nextLine();

        TopologicalSort ts = new TopologicalSort(V);
        for (int i = 0; i < E; i++) {
            int[] e = Arrays.stream(scanner.nextLine().split(" ")).mapToInt(Integer::parseInt).toArray();
            ts.addEdge(e[0], e[1]);
        }

        List<Integer> sortedNodes = ts.run();
        for (int n: sortedNodes) {
            System.out.println(n);
        }
    }
}


class TopologicalSort {

    private int size;
    private int[] inDeg;
    public Map<Integer, Set<Edge>> edges;

    public TopologicalSort(int size) {
        this.size = size;
        inDeg = new int[size];
        edges = new HashMap<>();
    }

    public void addEdge(int from, int to) {
        inDeg[to]++;
        if (!edges.containsKey(from)) {
            edges.put(from, new HashSet<>());
        }
        edges.get(from).add(new Edge(from, to, 1));
    }

    public List<Integer> run() {
        Stack<Integer> stack = new Stack<>();
        for (int i = 0; i < size; i++) {
            if (inDeg[i] == 0) {
                stack.push(i);
            }
        }

        List<Integer> sortedNodes = new ArrayList<>();
        while (!stack.isEmpty()) {
            int node = stack.pop();
            sortedNodes.add(node);
            for (Edge e: edges.getOrDefault(node, Collections.emptySet())) {
                inDeg[e.to]--;
                if (inDeg[e.to] == 0) {
                    stack.push(e.to);
                }
            }
        }

        // Not DAG
        if (sortedNodes.size() < size) {
            return Collections.emptyList();
        }
        return sortedNodes;
    }
}


class Edge {
    public final int from;
    public final int to;
    public final int cost;

    public Edge(int from, int to, int cost) {
        this.from = from;
        this.to = to;
        this.cost = cost;
    }

    @Override
    public boolean equals(Object o) {
        if (this == o) return true;
        if (o == null || getClass() != o.getClass()) return false;
        Edge edge = (Edge) o;
        return from == edge.from &&
                to == edge.to &&
                cost == edge.cost;
    }

    @Override
    public int hashCode() {
        return Objects.hash(from, to, cost);
    }
}