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<string>
#include<iomanip>
#include<cmath>
#include<vector>
#include<algorithm>
#include<queue>

using namespace std;

#define int long long
#define rep(i,n) for(int i = 0; i < (n); i++)
#define INF ((long long)1e18)
#define EPS (1e-5)
#define MOD ((int)1e9+7)
#define endl "\n"

#define yn(f) ((f)?"Yes":"No")
#define YN(f) ((f)?"YES":"NO")

#define MAX 110000

int N, M;
int root;

vector<int> G[MAX];
int nroot[MAX];
int con[MAX];

int parent[MAX];

void solve(){
	queue<int> Q;
	
	Q.push(root);
	
	while(!Q.empty()){
		int temp = Q.front(); Q.pop();
		for(int x : G[temp]){
			con[x]++;
			if(con[x] != nroot[x]) continue;
			Q.push(x);
			parent[x] = temp;
		}
	}
}

signed main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	cout<<fixed<<setprecision(7);
	
	int A, B;
	
	cin>>N>>M;
	
	for(int i = 0; i < N+M-1; i++){
		cin>>A>>B;
		G[A].push_back(B);
		nroot[B]++;
	}
	
	for(int i = 1; i <= N; i++){
		if(!nroot[i]){
			root = i;
			break;
		}
	}
	
	solve();
	
	for(int i = 1; i <= N; i++){
		cout<<parent[i]<<endl;
	}
	
	return 0;
}