Problem:
Score : 500 points

Problem Statement

There is a rooted tree (see Notes) with N vertices numbered 1 to N.
Each of the vertices, except the root, has a directed edge coming from its parent.
Note that the root may not be Vertex 1.

Takahashi has added M new directed edges to this graph.
Each of these M edges, u \rightarrow v, extends from some vertex u to its descendant v.

You are given the directed graph with N vertices and N-1+M edges after Takahashi added edges.
More specifically, you are given N-1+M pairs of integers, (A_1, B_1), ..., (A_{N-1+M}, B_{N-1+M}), which represent that the i-th edge extends from Vertex A_i to Vertex B_i.

Restore the original rooted tree.

Notes

For "tree" and other related terms in graph theory, see the article in Wikipedia, for example.

Constraints

3 \leq N

1 \leq M

N + M \leq 10^5

1 \leq A_i, B_i \leq N

A_i \neq B_i

If i \neq j, (A_i, B_i) \neq (A_j, B_j).

The graph in input can be obtained by adding M edges satisfying the condition in the problem statement to a rooted tree with N vertices.

Input

Input is given from Standard Input in the following format:

N M
A_1 B_1
:
A_{N-1+M} B_{N-1+M}

Output

Print N lines.
In the i-th line, print 0 if Vertex i is the root of the original tree, and otherwise print the integer representing the parent of Vertex i in the original tree.

Note that it can be shown that the original tree is uniquely determined.

Sample Input 1

3 1
1 2
1 3
2 3

Sample Output 1

0
1
2

The graph in this input is shown below:

It can be seen that this graph is obtained by adding the edge 1 \rightarrow 3 to the rooted tree 1 \rightarrow 2 \rightarrow 3.

Sample Input 2

6 3
2 1
2 3
4 1
4 2
6 1
2 6
4 6
6 5

Sample Output 2

6
4
2
0
6
2

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define int long long

signed main(){
       int n,m;
       cin >>n>>m;
       vector<int> tk[n+1],tr[n+1];
       int a,b;
         stack<int> que;
       
    for(int i=0;i<n+m-1;i++){
           cin >> a>>b;
           tk[b].push_back(a);
           tr[a].push_back(b);
       }
        int u=1;
        while(!tk[u].empty())u=tk[u][0];
    int ans2[n+1]={};
    int ans[n+1]={};
  
    ans2[u]=0;
    for(int i=0;i<tr[u].size();i++){
        
        
        ans2[tr[u][i]]=u;
        ans[tr[u][i]]=1;
        que.push(tr[u][i]);
    }
    while(!que.empty()){
        u=que.top();
        que.pop();
        for(int i=0;i<tr[u].size();i++){
        
     //   ans[tr[u][i]]=ans[u]+1;
        if(ans[tr[u][i]]==0){
           ans2[tr[u][i]]=u;
           ans[tr[u][i]]=ans[u]+1;
        que.push(tr[u][i]); 
        }else{
            if(ans[tr[u][i]]<ans[u]+1)
                ans2[tr[u][i]]=u;
                ans[tr[u][i]]=ans[u]+1;
        }
        
       
        }

    }
    for(int i=1;i<n+1;i++){
        cout << ans2[i]<<endl;
    }
    
    return 0;
}