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 <vector>
#include <queue>
using namespace std;


vector<int> V[100000];

int findRoot( vector<int>& B, int N, int M ){
    vector<bool> isAppeared(N, false);  // B に現れてないのが root
    for( int i = 0; i < N-1+M; ++i ) isAppeared[B[i]] = true;
    for( int i = 0; i < N; ++i ){
        if( !isAppeared[i] ) return i;
    }
    return -1;
}


int main()
{
    int N, M;
    cin >> N >> M;
    vector<int> A(N-1+M), B(N-1+M);
    for( int i = 0; i < N+M-1; ++i ){
        cin >> A[i] >> B[i];
        --A[i], --B[i];
        V[A[i]].push_back( B[i] );
    }


    int root = findRoot(B, N, M);
    
    vector<int> dist(N), parent(N);
    queue<int> s;
    s.push( root );
    while( !s.empty() )
    {
        int tmp = s.front();
        s.pop();
        auto edge = V[tmp].begin();
        while( edge != V[tmp].end() ){
            if( dist[*edge] <= dist[tmp]+1 ){
                dist[*edge] = dist[tmp]+1; 
                parent[*edge] = tmp+1;
                s.push( *edge );
                ++edge;
            }else{
                V[tmp].erase( edge );
            }
        }
    }
    vector<int> tmp(3);
    tmp[1] = 1, tmp[2] = 2;
    auto itr = tmp.begin();
    while( itr != tmp.end() ){
        cout << *itr << endl;
        tmp.erase( itr );
    }
    for( int i = 0; i < N; ++i ) cout << parent[i] << endl;

    return 0;
}