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 <iostream>
#include <string>
#include <cmath>
#include<algorithm>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<iomanip>
#define _USE_MATH_DEFINES
#include <math.h>
#include <functional>
#include<complex>
using namespace std;

#define rep(i,x) for(ll i=0;i<x;i++)
#define repn(i,x) for(ll i=1;i<=x;i++)

typedef long long ll;
const ll INF = 1e17;
const ll MOD = 1000000007;
const ll MAX = 4000001;
const double eps = 1E-18;

ll max(ll a, ll b) {
	if (a > b) { return a; }
	return b;
}

ll min(ll a, ll b) {
	if (a > b) { return b; }
	return a;
}

ll gcd(ll a, ll b) {
	if (b == 0) { return a; }
	if (a < b) { return gcd(b, a); }
	return gcd(b, a % b);
}

ll lcm(ll a, ll b) {
	return a / gcd(a, b) * b;
}

struct edge {
	ll ind;
	ll fr;
	ll to;
	ll d;
};

class mint {
	long long x;
public:
	mint(long long x = 0) : x((x% MOD + MOD) % MOD) {}
	mint operator-() const {
		return mint(-x);
	}
	mint& operator+=(const mint& a) {
		if ((x += a.x) >= MOD) x -= MOD;
		return *this;
	}
	mint& operator-=(const mint& a) {
		if ((x += MOD - a.x) >= MOD) x -= MOD;
		return *this;
	}
	mint& operator*=(const  mint& a) {
		(x *= a.x) %= MOD;
		return *this;
	}
	mint operator+(const mint& a) const {
		mint res(*this);
		return res += a;
	}
	mint operator-(const mint& a) const {
		mint res(*this);
		return res -= a;
	}
	mint operator*(const mint& a) const {
		mint res(*this);
		return res *= a;
	}

	mint operator^(const ll& t)const {
		if (!t) return 1;
		mint a = pow(t >> 1);
		a *= a;
		if (t & 1) a *= *this;
		return a;
	}

	mint pow(ll t) const {
		if (!t) return 1;
		mint a = pow(t >> 1);
		a *= a;
		if (t & 1) a *= *this;
		return a;
	}
	// for prime MOD
	mint inv() const {
		return pow(MOD - 2);
	}
	mint& operator/=(const mint& a) {
		return (*this) *= a.inv();
	}
	mint operator/(const mint& a) const {
		mint res(*this);
		return res /= a;
	}

	friend ostream& operator<<(ostream& os, const mint& m) {
		os << m.x;
		return os;
	}
};

///////////////////////////
struct bellmanford {
	ll N;//頂点数(1-index)
	vector<vector<edge>> g;//有向グラフ
	ll s;


	vector<edge> e;
	vector<ll> ans;//頂点sからiまでの距離(負閉路がない場合)
	ll minuscycle;//sから到達可能な負閉路があるか

	bellmanford(ll N0, ll s0, vector<vector<edge>> g0) {
		N = N0;
		s = s0;
		g = g0;

		ans.assign(N + 1, INF);
		ans[s] = 0;

		repn(i, N) {
			for (edge ed : g[i]) {
				e.push_back(ed);
			}
		}

		minuscycle = 0;

		repn(i, N) {
			for (edge ed : e) {
				if (ans[ed.fr] != INF && ans[ed.fr] + ed.d < ans[ed.to]) {
					ans[ed.to] = ans[ed.fr] + ed.d;
					if (i == N) { minuscycle = 1; }
				}
			}
		}

	}
};


int main() {
	ll N, M;
	cin >> N >> M;

	vector<vector<edge>> g(N + 1);
	vector<bool> root(N + 1,1);

	repn(i, N + M - 1) {
		ll a, b;
		cin >> a >> b;
		g[a].push_back({i,a,b,-1});
		root[b] = 0;
	}

	ll rt = 0;
	repn(i, N) {
		if (root[i]) { rt = i; }
	}

	bellmanford dij(N, rt, g);

	
	vector<ll> ans(N + 1, 0);
	repn(v, N) {
		for (edge e : g[v]) {
			ll w = e.to;
			if (dij.ans[w] == dij.ans[v] - 1) { ans[w] = v; }
		}
	}

	repn(i, N) {
		cout << ans[i] << endl;
	}

	
	



	
	

	
	
	system("PAUSE");

}