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>
#define INF 1e9
#define llINF 1e18
#define MOD 1000000007
#define pb push_back
#define mp make_pair 
#define F first
#define S second
#define ll long long
#define vi vector<int>
#define vvi vector<vi>
#define substr(s,f,t) (s.substr(f-1,t-f+1))
#define ALL(a) (a).begin(),(a).end()
#define Yes(hoge) cout<<((hoge)?"Yes":"No")<<endl;
#define YES(hoge) cout<<((hoge)?"YES":"NO")<<endl;
using namespace std;
struct Grid{ll x,y,t;};
struct Edge{ll to,cost;};
struct Graph{vector<vector<Edge>>E;int V;
  const ll Inf = llINF;const int MAX_V=201010;vector<ll>d;
  Graph(int n):E(n){d.resize(MAX_V);E.resize(n);V=n;}
  void init(){for(int i=0;i<MAX_V;i++)d[i]=Inf;}
  void add_edge(ll from,ll to,ll cost){E[from].pb({to,cost});}
};
vvi E(100010);
int par[100010];
int dist[100010];
int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  int n,m;cin>>n>>m;
  vi hoge(100010);
  for(int i=0;i<n+m-1;i++){
    int a,b;cin>>a>>b;
    E[a].pb(b);
    hoge[b]++;
  }
  for(int i=1;i<=n;i++){
    if(hoge[i]!=0)continue;
    queue<pair<ll,ll>>que;
    que.push(mp(i,0));
    while(!que.empty()){
      pair<int,int>now=que.front();
      que.pop();
      for(int j=0;j<E[now.F].size();j++){
	if(dist[E[now.F][j]]){
	  par[E[now.F][j]]=now.F;
	  dist[E[now.F][j]]=now.S+1;
	}else{par[E[now.F][j]]=now.F;
	  dist[E[now.F][j]]=now.S+1;
	  que.push(mp(E[now.F][j],now.S+1));}
	
      }
    }
  
 }
for(int i=0;i<n;i++)cout<<par[i]<<endl;
return 0;
}