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;
typedef long long ll;


ll n, m;
vector<ll> depth(10000000, 0);
vector<vector<ll> > chil(1000000);
vector<ll> ra(1000000, 0);
vector<ll> parent(1000000, -1);

void dfs(ll x, ll d) {
	for (int i = 0; i < chil[x].size(); ++i)
	{
		ll c = chil[x][i];
		if (depth[c] < d+1)
		{
			depth[c] = d+1;
			parent[c] = x;
			dfs(c, d+1);
		}
	}
	return;
}

int main() {
	cin >> n >> m;
	for (int i = 0; i < n+m-1; ++i)
	{
		ll a, b;
		cin >> a >> b;
		chil[a-1].push_back(b-1);
		ra[b-1] += 1;
	}
	ll par;
	for (int i = 0; i < n; ++i)
	{
		if (ra[i] == 0)
		{
			par = i;
			break;
		}
	}
	parent[par] = -1;
	dfs(par, 0);
	for (int i = 0; i < n; ++i)
	{
		cout << parent[i]+1 << "\n";
	}
	return 0;
}