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.util.*;
import java.io.*;

public class CFE_25_E {
	FastScanner in;
	PrintWriter out;

	static int n, m;
	static ArrayList<Integer> a[], ar[], top;
	static boolean v[];

	public void solve() {
		n = in.nextInt();
		m = in.nextInt();

		v = new boolean[n];
		a = new ArrayList[n];
		ar = new ArrayList[n];
		for (int i = 0; i < n; i++) {
			a[i] = new ArrayList<Integer>();
			ar[i] = new ArrayList<Integer>();
		}
		top = new ArrayList<Integer>();
		int deg[] = new int[n];

		for (int i = 0; i < m; i++) {
			int f, t;
			a[(f = in.nextInt() - 1)].add((t = in.nextInt() - 1));
			ar[t].add(f);
			deg[f]++;
		}

		PriorityQueue<Integer> q = new PriorityQueue<Integer>(Collections.reverseOrder());

		for (int i = 0; i < n; i++) {
			if (deg[i] == 0) {
				q.add(i);
			}
		}

		int vis = 0;

		int res[] = new int[n];
		int cur = n;
		for (int x : top) {
			res[x] = cur++;
		}

		while (!q.isEmpty()) {
			int v = q.poll();
			res[v] = cur--;
			for (int z : ar[v]) {
				if (--deg[z] == 0) {
					q.add(z);
				}
			}
		}

		for (int x : res) {
			out.print(x + " ");
		}
	}

	public void DFS(int x) {
		v[x] = true;
		for (int z : a[x]) {
			if (!v[z]) {
				DFS(z);
			}
		}
		top.add(x);
	}

	public void run() {
		in = new FastScanner();
		out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		solve();
		out.close();
	}

	class FastScanner {
		BufferedReader br;
		StringTokenizer st;

		FastScanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			while (st == null || !st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

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

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

	public static void main(String[] args) {
		new CFE_25_E().run();
	}
}
