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<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iomanip>
#include<vector>
#include<queue>
#include<deque>
#include<string>
#include<utility>
#include<cmath>
#include<algorithm>
#include<cctype>
#include<set>
#include<map>
#include<bitset>
#include<stack>
#include<ctime>
#include<cassert>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<ll,int> pli;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<pii> vpii;
typedef vector<pli> vpli;

inline ll read(){
    ll Hashimoto=0;
	bool Kanna=1;
    char I_Love=getchar();
    while(I_Love<'0'||I_Love>'9'){
        if(I_Love=='-')Kanna=0;
        I_Love=getchar();
    }
    while(I_Love>='0'&&I_Love<='9'){
        Hashimoto=Hashimoto*10+I_Love-'0';
        I_Love=getchar();
    }
    return (Kanna?Hashimoto:-Hashimoto);
}
template<typename T1,typename T2> inline void Umax(T1 &a,T2 b){
	if(a<b)a=b;
}
template<typename T1,typename T2> inline void Umin(T1 &a,T2 b){
	if(a>b)a=b;
}

#define I_Love_Hashimoto_Kanna main
#define fastio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define filein(s) freopen(s,"r",stdin);
#define fileout(s) freopen(s,"w",stdout);
#define file freopen("I_Love_Hashimoto_Kanna.out","w",stdout);
#define RE cout<<"I_Love_Hashimoto_Kanna"<<endl;
#define Tone(Kanna) cout<<Kanna<<endl;
#define Tall(Hashimoto,Kanna) for(int I_Love=0;I_Love<(Kanna);++I_Love)cout<<Hashimoto[I_Love]<<(I_Love==(Kanna)-1?"\n":" ");
#define FOR(I_Love,Hashimoto,Kanna) for(int I_Love=Hashimoto;I_Love<(Kanna);++I_Love)
#define MP(Hashimoto,Kanna) make_pair(Hashimoto,Kanna)
#define REV(Kanna) reverse(Kanna.begin(),Kanna.end());
#define SORT(Kanna) sort(Kanna.begin(),Kanna.end());
#define UNIQUE(Kanna) Kanna.erase(unique(Kanna.begin(),Kanna.end()),Kanna.end());
#define inf (1000000000)
#define linf (1000000000000000000ll)
#define mod (1000000007)

int n,m;
int indegree[111111];
vi g[111111];
queue<int>q;
bool v[111111];
int ans[111111];

int I_Love_Hashimoto_Kanna() {
	//完全想清楚了再开始写。
	//写不顺、不知道怎么写、很乱的时候，停下来好好想想。
	//做得慢总比做不出好。
	fastio;
	n=read();m=read();
	FOR(i,1,n+m){
		int a=read()-1,b=read()-1;
		g[a].push_back(b);
		indegree[b]++;
	}
	int root=-1;
	FOR(i,0,n){
		if(indegree[i]==0)root=i;
	}
	assert(root!=-1);
	ans[root]=-1;
	v[root]=1;
	q.push(root);
	while(!q.empty()){
		int x=q.front();
		q.pop();
		FOR(i,0,g[x].size()){
			int to=g[x][i];
			ans[to]=x;
			if(v[to])continue;
			q.push(to);
			v[to]=1;
		}
	}
	FOR(i,0,n){
		printf("%d\n",ans[i]+1);
	}
	
	
	return 0;
}