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.BufferedReader;
import java.io.DataInputStream;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

import javax.swing.plaf.synth.SynthSpinnerUI;
import javax.swing.text.AbstractDocument.LeafElement;


public class Main2 {
	static List<List<Integer>> list = new ArrayList<List<Integer>>();
	static List<List<Integer>> reverse = new ArrayList<List<Integer>>();
	static int[] pre;
	static boolean[] visited;
	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int m = sc.nextInt();
		visited = new boolean[n];
		pre = new int[n];
		for (int i = 0; i < n; i++) {
			list.add(new ArrayList<Integer>());
		}
		for (int i = 0; i < m; i++) {
			int from = sc.nextInt() -1;
			int to = sc.nextInt() - 1;
			list.get(to).add(from);
		}
		
		int[] inDegree = new int[n];
		int[] outDegree = new int[n];
		for (int i = 0; i < outDegree.length; i++) {
			outDegree[i] = list.get(i).size();
		}
		for(int i = 0; i < n; i++) {
			for(int v : list.get(i)) {
				inDegree[v]++;
			}
			
		}
		int pointer = n;

		PriorityQueue<Integer> pq = new PriorityQueue<Integer>(); 
		for(int i = 0; i < n; i++) {
			if(inDegree[i] == 0) {
				pq.add(n-i);
			}
		}
		int[] label = new int[n];
		while(!pq.isEmpty()) {
			int next = pq.poll();
			label[n-next] = pointer--;
			for(int v : list.get(n-next)) {
				inDegree[v]--;
				if (inDegree[v] == 0) {
					pq.add(n-v);
				}
			}
		}
		for(int i = 0; i < n; i++) {
			System.out.print(label[i] + " ");
		}
		
	}

}

