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: Accepted)

#include <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <string>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <limits.h>
#include <math.h>
#include <functional>
#include <bitset>
#include <iomanip>
#include <cassert>

#define repeat(i,n) for (long long i = 0; (i) < (n); ++ (i))
#define debug(x) cerr << #x << ": " << x << '\n'
#define debugArray(x,n) for(long long i = 0; (i) < (n); ++ (i)) cerr << #x << "[" << i << "]: " << x[i] << '\n'
#define debugArrayP(x,n) for(long long i = 0; (i) < (n); ++ (i)) cerr << #x << "[" << i << "]: " << x[i].first<< " " << x[i].second << '\n'

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> Pii;
typedef vector<int> vint;
typedef vector<ll> vll;
const ll INF = LLONG_MAX;
const ll MOD = 1e9+7;

bool visit(const vector<vint> &g, int v, vector<int> &order, vector<int> &color) {
  color[v] = 1;
  for(auto e: g[v]) {
    if (color[e] == 2) continue;
    if (color[e] == 1) return false;
    if (!visit(g, e, order, color)) return false;
  }
  order.push_back(v); color[v] = 2;
  return true;
}
bool topologicalSort(const vector<vint> &g, vector<int> &order) {
  int n = g.size();
  vector<int> color(n);
  repeat(u, n) if (!color[u] && !visit(g, u, order, color))
    return false;
  reverse(order.begin(),order.end());
  return true;
}

int N,M;
vector<vint> G;
vint par;
int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  cin>>N>>M;
  G.resize(N);
  int root=0;
  repeat(i,N-1+M){
    int A,B;cin>>A>>B;A--;B--;
    G[A].push_back(B);
    if(root==B){
      root=A;
    }
  }
  //debug(root);
  vint order;
  topologicalSort(G,order);

  par.assign(N,0);
  repeat(i,N){
    int v=order[i];
    for(int u:G[v]){
      par[u]=v+1;
    }
  }
  repeat(i,N){
    cout << par[i] << endl;
  }
  return 0;
}