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<iostream>
#include<stdio.h>
#include<ext/rope>
#include<vector>
using namespace std;
int p[300010],nex[300010],head[300010],e;
void add(int a,int b){
    p[e]=b;
    nex[e]=head[a];
    head[a]=e++;
}
int vis[300010],A[300010],B[300010],k;
int dfs(int u){
    vis[u]=-1;//cout<<6<<endl;
    int i,v;
    for(i=head[u];i;i=nex[i]){
        v=p[i];
        if(vis[v]==-1) return 0;
        if(vis[v]==0&&dfs(v)==0) return 0;
    }
    vis[u]=1;
    B[u]=k;
    A[k--]=u;
    return 1;
}

int fa[300010];
vector<int>g[300010];
int main(){
    int j,i,n,m,v,a,b;
    cin>>n>>m;
    k=n;
    e=1;
    for(i=1;i<n+m;i++){
        cin>>a>>b;
        add(a,b);
        g[b].push_back(a);
    }
    for(i=1;i<=n;i++) if(vis[i]==0) dfs(i);
    for(i=1;i<=n;i++){
        for(j=0;j<g[i].size();j++){
            v=g[i][j];
            if(fa[i]==0||B[v]>B[fa[i]]) fa[i]=v;
        }
    }
   // for(i=1;i<=n;i++) cout<<A[i]<<endl;

    //cout<<A[1]<<endl;
    for(i=1;i<=n;i++) cout<<fa[i]<<endl;
    //out<<fa[n]<<endl;
    return 0;


}