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 <bits/stdc++.h>
using namespace std;
struct Fast {Fast(){std::cin.tie(0);ios::sync_with_stdio(false);cout.precision(20);}} fast;

/* define */
#define FOR(I,X,Y) for(long long (I)=(X);(I)<(Y);(I)++)
#define REP(I,X,Y) for(long long (I)=(Y)-1;(I)>=(X);(I)--)
#define ALL(X) (X).begin(),(X).end()
#define pb push_back
#define COUNT(V,X) (upper_bound((V).begin(),(V).end(),X)-lower_bound((V).begin(),(V).end(),X))
#define debug(x) cerr<<#x<<':'<<x<<endl;
#define DEBUG(v) cerr<<#v<<':';for(auto xXx:v)cerr<<xXx<<' ';cerr<<endl;
#define Yes(X) if(X){cout<<"Yes"<<endl;}else{cout<<"No"<<endl;}
#define YES(X) if(X){cout<<"YES"<<endl;}else{cout<<"NO"<<endl;}
#define ctoi(C) (C-'0')

/* alias */
using ll = long long;
using ld = long double;
using vi = vector<int>;
using vii = vector<vector<int>>;
using vl = vector<long long>;
using vll = vector<vector<long long>>;
using pi = pair<int,int>;
using pl = pair<long long,long long>;
template<typename T> using PQ = priority_queue<T>;
template<typename T> using minPQ = priority_queue<T, vector<T>, greater<T>>;

/* const */
const long long dx[] = {1,0,-1,0};
const long long dy[] = {0,1,0,-1};
const long long dx8[] = {1,1,0,-1,-1,-1,0,1};
const long long dy8[] = {0,1,1,1,0,-1,-1,-1};
const long long dx9[] = {1,1,0,-1,-1,-1,0,1,0};
const long long dy9[] = {0,1,1,1,0,-1,-1,-1,0};
const int INF = 1000000007;
const long long LINF = 1000000000000000007;

/* func */
template <typename T> inline bool chmin(T& a, const T& b) {if (a > b) a = b; return a > b;}
template <typename T> inline bool chmax(T& a, const T& b) {if (a < b) a = b; return a < b;}
long long max(long long x,int y){return max(x,(long long)y);}
long long max(int x,long long y){return max((long long)x,y);}
long long min(long long x,int y){return min(x,(long long)y);}
long long min(int x,long long y){return min((long long)x,y);}

/* liblary */



/* main */

ll N,M;
vector<vector<ll>> adj;
map<ll,ll> depth;

ll func(ll n){
	if(depth[n])return depth[n];
	ll ans = 1;
	for(auto x:adj[n])chmax(ans,func(x)+1);
	return depth[n] = ans;
}

signed main(){
	cin >> N >> M;
	adj.resize(N);
	FOR(i,0,N-1+M){
		ll A,B;
		cin >> A >> B;
		A--;
		B--;
		adj[B].pb(A);
	}
	FOR(i,0,N)func(i);
	FOR(i,0,N){
		bool flag = 1;
		for(auto x:adj[i]){
			if(depth[x] + 1 == depth[i]){
				cout << x+1 << endl;
				flag = 0;
				break;
			}
		}
		if(flag)cout << 0 << endl;
	}
}