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.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.StringTokenizer;




public class solver {
    BufferedReader in;
    PrintWriter out;
    StringTokenizer tok;
    
    String readString() throws IOException{
        while (tok==null || !tok.hasMoreTokens()){
            try{
                tok=new StringTokenizer(in.readLine());             
            }
            catch(Exception e){
                return null;
            }            
        }
        return tok.nextToken();
    }
    int readInt() throws NumberFormatException, IOException{
        return Integer.parseInt(readString());
    }
    long readLong() throws NumberFormatException, IOException{
        return Long.parseLong(readString());
    }
    double readDouble() throws NumberFormatException, IOException{
        return Double.parseDouble(readString());
    }
    
    final static boolean OJ=System.getProperty("ONLINE_JUDGE")!=null;
    
    
    void init() throws FileNotFoundException{        
        if (OJ){
            in=new BufferedReader(new InputStreamReader(System.in));
            out=new PrintWriter(System.out);  
        }else{          
            in=new BufferedReader(new FileReader("input.txt"));  
            out=new PrintWriter("output.txt");  
            
            in=new BufferedReader(new InputStreamReader(System.in));
            out=new PrintWriter(System.out);
        }
    }    
    int[] iMas(int n) throws NumberFormatException, IOException{
        int[] a=new int[n];
        for (int i=0;i<n;i++){
            a[i]=readInt();
        }
        return a;
    }
    
    
    int gcd (int a, int b) {
        if (b == 0)
            return a;
        else
            return gcd (b, a % b);
    }   
    
    
    
    class Pair implements Comparable<Pair>{
        int b;
        int d;
        String s;
        
        Pair(int x,int y,String s){
            b=x;
            d=y;            
            this.s=s;
        }
                
        @Override
        public int compareTo(Pair o) {          
            return b-o.b;
        }                       
    }
    
    int get(int i,int j,int n){
        if (i+j<=n) return i+j;
        return i+j-n;
    }
    
    
    ArrayList<Integer>[] graph;
    boolean[] vis;
    ArrayList<Integer> topsort;
    
    
    void dfs(int x){
        vis[x]=true;
        
        for (int y:graph[x]){
            if (!vis[y]) dfs(y);
        }
        topsort.add(x+1);
        
    }
    
    
    void solve() throws IOException{
        
        int n=readInt();
        int m=readInt();
        graph=new ArrayList[n];
        vis=new boolean[n];
        topsort=new ArrayList<Integer>();
        
        for (int i=0;i<n;i++){
            graph[i]=new ArrayList<Integer>();
        }
        
        for (int i=0;i<m;i++){
            int x=readInt()-1;
            int y=readInt()-1;
            graph[x].add(y);
        }
        for (int i=0;i<n;i++){
            if (!vis[i]) dfs(i);
        }
        for (int i:topsort) out.print(i+" ");
        
        
        
    }
    
    
    void run() throws NumberFormatException, IOException{
        init();
         long h=System.currentTimeMillis();        
        solve();
         System.err.println(System.currentTimeMillis()-h);
        out.close();
    }
    
    
    

    
    public static void main(String[] args) throws NumberFormatException, IOException {
        new solver().run();
    }

}