Problem: The employees of the R1 company often spend time together: they watch football, they go camping, they solve contests. So, it's no big deal that sometimes someone pays for someone else.

Today is the day of giving out money rewards. The R1 company CEO will invite employees into his office one by one, rewarding each one for the hard work this month. The CEO knows who owes money to whom. And he also understands that if he invites person x to his office for a reward, and then immediately invite person y, who has lent some money to person x, then they can meet. Of course, in such a situation, the joy of person x from his brand new money reward will be much less. Therefore, the R1 CEO decided to invite the staff in such an order that the described situation will not happen for any pair of employees invited one after another.

However, there are a lot of employees in the company, and the CEO doesn't have a lot of time. Therefore, the task has been assigned to you. Given the debt relationships between all the employees, determine in which order they should be invited to the office of the R1 company CEO, or determine that the described order does not exist.

Input

The first line contains space-separated integers n and m <image> — the number of employees in R1 and the number of debt relations. Each of the following m lines contains two space-separated integers ai, bi (1 ≤ ai, bi ≤ n; ai ≠ bi), these integers indicate that the person number ai owes money to a person a number bi. Assume that all the employees are numbered from 1 to n.

It is guaranteed that each pair of people p, q is mentioned in the input data at most once. In particular, the input data will not contain pairs p, q and q, p simultaneously.

Output

Print -1 if the described order does not exist. Otherwise, print the permutation of n distinct integers. The first number should denote the number of the person who goes to the CEO office first, the second number denote the person who goes second and so on.

If there are multiple correct orders, you are allowed to print any of them.

Examples

Input

2 1
1 2


Output

2 1 


Input

3 3
1 2
2 3
3 1


Output

2 1 3 

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.FileOutputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.List;
import java.util.PriorityQueue;
import java.util.Random;
import java.util.Scanner;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.TreeSet;
import java.lang.Long;
import java.math.BigDecimal;
import java.math.BigInteger;

/**
 * Problem statement: http://codeforces.com/problemset/problem/431/D
 * 
 * @author thepham
 * 
 */

public class D_CodeStrike2014_Round1 {
	static long Mod = 1000000007;

	public static void main(String[] args) throws FileNotFoundException {
		// PrintWriter out = new PrintWriter(new FileOutputStream(new File(
		// "output.txt")));
		PrintWriter out = new PrintWriter(System.out);
		Scanner in = new Scanner();

		int n = in.nextInt();
		int m = in.nextInt();
		HashSet<Integer>[] map = new HashSet[n];
		HashSet<Integer>[] reverse = new HashSet[n];
		for (int i = 0; i < n; i++) {
			map[i] = new HashSet();
			reverse[i] = new HashSet();
		}
		int[] depend = new int[n];
		int[] point = new int[n];
		for (int i = 0; i < m; i++) {
			int a = in.nextInt() - 1;
			int b = in.nextInt() - 1;
			map[a].add(b);
			reverse[b].add(a);
			depend[a]++;
			point[b]++;
		}

		boolean ok = true;
		int last = -1;
		int[] result = new int[n];

		boolean[] check = new boolean[n];
		TreeSet<Node> tree = new TreeSet();
		for (int i = 0; i < n; i++) {
			tree.add(new Node(i, point[i], depend[i]));
		}
		for (int i = 0; i < n; i++) {
			int index = -1;
			LinkedList<Node> q=  new LinkedList();
			while(!tree.isEmpty()){
				Node node = tree.pollFirst();
				if(last != -1 && map[last].contains(node.index)){
					q.add(node);
					
				}else{
					index = node.index;
					break;
				}
			}

			if (index == -1) {
				ok = false;
				break;
			}
			tree.addAll(q);
			// System.out.println(index);
			check[index] = true;
			result[i] = index;
			last = index;
			for (int j : map[index]) {
				if(check[j]){
					continue;
				}
				tree.remove(new Node(j, point[j], depend[j]));
				point[j]--;
				tree.add(new Node(j,point[j] , depend[j]));
			}
			for (int j : reverse[index]) {
				if(check[j]){
					continue;
				}
				tree.remove(new Node(j, point[j], depend[j]));
				depend[j]--;
				tree.add(new Node(j,point[j] , depend[j]));
			}
		}
		if (!ok) {
			out.println(-1);
		} else {
			for (int i : result) {
				out.print((i + 1) + " ");
			}
		}

		out.close();
	}

	static class Node implements Comparable<Node> {
		int index, num, other;

		public Node(int index, int num, int other) {
			this.index = index;
			this.num = num;
			this.other = other;
		}

		@Override
		public int compareTo(Node o) {
			if (other != o.other) {
				return other - o.other;
			}
			if (num != o.num){
				return o.num - num;
			}
			return index - o.index;
		}

	}

	// public static long cal(int k, long cur, long[][]p){
	//
	// }

	public static int[] buildKMP(String val) {
		int[] data = new int[val.length() + 1];
		int i = 0;
		int j = -1;
		data[0] = -1;
		while (i < val.length()) {
			while (j >= 0 && val.charAt(i) != val.charAt(j)) {
				j = data[j];
			}
			i++;
			j++;

			data[i] = j;
			// System.out.println(val + " " + i + " " + data[i]);
		}
		return data;
	}

	static int find(int index, int[] u) {
		if (u[index] != index) {
			return u[index] = find(u[index], u);
		}
		return index;
	}

	static int crossProduct(Point a, Point b) {
		return a.x * b.y + a.y * b.x;
	}

	static long squareDist(Point a) {
		long x = a.x;
		long y = a.y;
		return x * x + y * y;
	}

	static Point minus(Point a, Point b) {
		return new Point(a.x - b.x, a.y - b.y);
	}

	public static boolean nextPer(int[] data) {
		int i = data.length - 1;
		while (i > 0 && data[i] < data[i - 1]) {
			i--;
		}
		if (i == 0) {
			return false;
		}
		int j = data.length - 1;
		while (data[j] < data[i - 1]) {
			j--;
		}
		int temp = data[i - 1];
		data[i - 1] = data[j];
		data[j] = temp;
		Arrays.sort(data, i, data.length);
		return true;
	}

	static class Point {
		int x, y;

		public Point(int x, int y) {
			super();
			this.x = x;
			this.y = y;
		}

		public String toString() {
			return "{" + x + " " + y + "}";
		}

	}

	static long gcd(long a, long b) {
		if (b == 0) {
			return a;
		}
		return gcd(b, a % b);
	}

	public static class FT {

		int[] data;

		FT(int n) {
			data = new int[n];
		}

		public void update(int index, int value) {
			while (index < data.length) {
				data[index] += value;
				index += (index & (-index));
			}
		}

		public int get(int index) {
			int result = 0;
			while (index > 0) {
				result += data[index];
				index -= (index & (-index));
			}
			return result;

		}
	}

	static class Scanner {

		BufferedReader br;
		StringTokenizer st;

		public Scanner() throws FileNotFoundException {
			// System.setOut(new PrintStream(new
			// BufferedOutputStream(System.out), true));
			br = new BufferedReader(new InputStreamReader(System.in));
			// br = new BufferedReader(new InputStreamReader(new
			// FileInputStream(
			// new File("input.txt"))));
		}

		public String next() {

			while (st == null || !st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (Exception e) {
					throw new RuntimeException();
				}
			}
			return st.nextToken();
		}

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

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

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

		public String nextLine() {
			st = null;
			try {
				return br.readLine();
			} catch (Exception e) {
				throw new RuntimeException();
			}
		}

		public boolean endLine() {
			try {
				String next = br.readLine();
				while (next != null && next.trim().isEmpty()) {
					next = br.readLine();
				}
				if (next == null) {
					return true;
				}
				st = new StringTokenizer(next);
				return st.hasMoreTokens();
			} catch (Exception e) {
				throw new RuntimeException();
			}
		}
	}

}