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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;;
#define ll long long
#define REP(i, n) for(int i = 0;i < n;i++)
#define REPR(i, n) for(int i = n;i >= 0;i--)
#define FOR(i, m, n) for(int i = m;i < n;i++)
#define FORR(i, m, n) for(int i = m;i >= n;i--)
#define INF 1<<30
#define LINF 1LL<<62
#define all(x) (x).begin(), (x).end()
#define mp make_pair
#define pb push_back
const int MOD = 1000000007;
 
typedef pair<int, int> P;
typedef pair<ll, ll> LP;
typedef pair<int, P> PP;
typedef pair<ll, LP> LPP;

int main(){
  ios::sync_with_stdio(false);
  cin.tie(0);
  ll n,m;cin >> n >> m;
  vector<ll> edge[n];
  bool isr[n];
  REP(i,n){
      isr[i] = true;
  }
  REP(i,n+m-1){
      ll a,b;
      cin >> a >> b;
      a--;b--;
      edge[a].push_back(b);
      isr[b] = false;
  }
  ll root;
  REP(i,n){
      if(isr[i]) root = i;
  }
  ll dis[n],par[n];
  REP(i,n){
      dis[i] = 0;
      par[i] = 0;
  }
  stack<ll> pq;
  pq.push(root);
  while(pq.size()){
      ll now = pq.top();
      pq.pop();
      REP(i,edge[now].size()){
          ll to = edge[now][i];
          if(dis[now] + 1 > dis[to]){
              dis[to] = dis[now] + 1;
              par[to] = now;
              pq.push(to);
          }
      }
  }
  REP(i,n){
      if(i==root)cout << 0 << endl;
      else cout << par[i] +1 << endl;
  }


}