You are given a directed acyclic graph with n vertices and m edges. There are no self-loops or multiple edges between any pair of vertices. Graph can be disconnected.

You should assign labels to all vertices in such a way that:

  * Labels form a valid permutation of length n — an integer sequence such that each integer from 1 to n appears exactly once in it. 
  * If there exists an edge from vertex v to vertex u then labelv should be smaller than labelu. 
  * Permutation should be lexicographically smallest among all suitable. 



Find such sequence of labels to satisfy all the conditions.

Input

The first line contains two integer numbers n, m (2 ≤ n ≤ 105, 1 ≤ m ≤ 105).

Next m lines contain two integer numbers v and u (1 ≤ v, u ≤ n, v ≠ u) — edges of the graph. Edges are directed, graph doesn't contain loops or multiple edges.

Output

Print n numbers — lexicographically smallest correct permutation of labels of vertices.

Examples

Input

3 3
1 2
1 3
3 2


Output

1 3 2 


Input

4 5
3 1
4 1
2 3
3 4
2 4


Output

4 1 2 3 


Input

5 4
3 1
2 1
2 3
4 5


Output

3 1 2 4 5 
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.List;
import java.util.PriorityQueue;
import java.util.Collection;
import java.util.Scanner;
import java.util.Queue;
import java.util.Comparator;
import java.util.ArrayList;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author llamaoo7
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskE solver = new TaskE();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskE {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            int n = in.nextInt();
            int m = in.nextInt();
            List<List<Integer>> parents = new ArrayList<>(n + 1);
            for (int i = 0; i < n + 1; i++) parents.add(new ArrayList<>());
            int[] outbound = new int[n + 1];
            for (int i = 0; i < m; i++) {
                int u = in.nextInt();
                int v = in.nextInt();
                parents.get(v).add(u);
                outbound[u]++;
            }
            Queue<Integer> queue = new PriorityQueue<>(Comparator.reverseOrder());
            for (int i = n; i >= 1; i--) if (outbound[i] == 0) queue.add(i);
            int[] labels = new int[n + 1];
            int curLabel = n;
            while (!queue.isEmpty()) {
                int cur = queue.poll();
                labels[cur] = curLabel;
                curLabel--;
                for (int p : parents.get(cur)) {
                    if (--outbound[p] == 0) queue.add(p);
                }
            }
            for (int i = 1; i <= n; i++) {
                out.print(labels[i]);
                if (i != n) out.print(" ");
            }
        }

    }
}

