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>
using namespace std;
typedef long long int ll;
typedef pair<ll, ll> P;
typedef vector<ll> VI;
typedef vector<VI> VVI;
const ll MOD = 998244353;
const ll INF = 2147483647;
const ll LINF = 9223372036854775807;
#define REP(i, n) for (int i = 0; i < n; i++)
#define ALL(v) v.begin(), v.end()

VVI edge(114514, VI(0));
vector<P> ans(114514,{-1,-1});
void dfs(int v, int p, int d) {
	if(ans[v].second<d){
		int k=ans[v].first;
		ans[v].first=p+1;
		ans[v].second=d;
		if(k!=-1) return;
	}
	for (auto to : edge[v]) {
		if (to != p) dfs(to, v, d+1);
	}
}

int main(){
	int n, m; cin >> n >> m;
	int a, b; vector<bool> f(n-1+m,0);
	REP(i,n-1+m){
		cin >> a >> b;
		a--; b--;
		f[b]=1;
		edge[a].push_back(b);
	}
	int r;
	REP(i,n-1+m){
		if(!f[i]){
			r=i;
			break;
		}
	}
	dfs(r,-1,0);
	REP(i,n){
		cout << ans[i].first << endl;
	}
	return 0;
}