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 <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <iterator>
#include <limits>
#include <list>
#include <locale>
#include <map>
#include <memory>
#include <new>
#include <numeric>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <stdio.h>
#include <typeinfo>
#include <utility>
#include <valarray>
#include <vector>
#define rep(i,a,n) for(ll i =(a);i<(n);++i)
#define urep(i,a,n) for(ll i = (a);i>=(n);--i)
#define all(x) (x).begin(),(x).end()
#define INF 1e18
const int mod = 1e9 + 7;
typedef long long ll;
using namespace std;
ll dx[4] = { 1,-1,0,0 };
ll dy[4] = { 0,0,1,-1 };
ll N,M,X,Y,A,B,C,Q,K,R,W,H,P,L;
string S,T;
ll ans[101010];
ll x[101010];
ll y[101010];
ll a[101010];
ll b[101010];
ll c[101010];
ll t[101010];
ll p[101010];
ll n[101010];
ll gcd(ll a, ll b) {
	if (b == 0)return a;
	else return gcd(b, a%b);
}
struct Edge{
	ll to,cost;
	 Edge(int to, int cost) : to(to), cost(cost) {}
};
typedef vector<vector<Edge> > AdjList;
AdjList graph;
vector<ll> dist;
ll outnum[101010];
ll innum[101010];


int main() {
	cin>>N>>M;
	graph=AdjList(N+1);
	rep(i,0,N-1+M){
		ll a,b;
		cin>>a>>b;
		graph[a].push_back(Edge(b,1));
		outnum[a]++;
		innum[b]++;
	}
	ll tmp;
	rep(i,1,N+1){
		if(innum[i]==0){
			ans[i]=0;
			tmp=i;
			break;
		}
	}
	queue<ll> q;
	q.push(tmp);
	while(!q.empty()){
		ll tmp =q.front();

		q.pop();
		rep(i,0,graph[tmp].size()){
			innum[graph[tmp][i].to]--;
			if(innum[graph[tmp][i].to]==0){
				ans[graph[tmp][i].to]=tmp;
				q.push(graph[tmp][i].to);
			}
		}
	}
	rep(i,1,N+1){
		cout<<ans[i]<<endl;
	}
	return 0;
}