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;
#define ll long long int
#define rep(i,n) for( int i = 0; i < n; i++ )
#define rrep(i,n) for( int i = n; i >= 0; i-- )
#define REP(i,s,t) for( int i = s; i <= t; i++ )
#define RREP(i,s,t) for( int i = s; i >= t; i-- )
#define dump(x)  cerr << #x << " = " << (x) << endl;
#define INF 2000000000
#define mod 1000000007
#define INF2 1000000000000000000

int ans[100010];

vector<int> G[100010];
vector<int> P[100010];
set<int> visited;

void dfs(int s) {
    //cout << s << endl;
    visited.insert(s);
    rep(i, G[s].size()) {
        ans[G[s][i]] = s;
    }
    rep(i, G[s].size()) {
        auto vis = visited.find(G[s][i]);
        if (vis != visited.end()) continue;
        bool f = true;
        rep(j, P[G[s][i]].size()) {
            auto it = visited.find(P[G[s][i]][j]);
            if (it == visited.end()) {
                f = false;
            }
            //cout << P[G[s][i]].size() << " " << P[G[s][i]][j] << "<-" << G[s][i]  << " " << f<<  endl;
            if(f == false) break;
        }
        //dump(f);
        if(f) dfs(G[s][i]);
    }
}


int main(void)
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    int N, M;
    cin >> N >> M;
    bool child[100010];
    rep(i, N + M - 1) {
        int s, t;
        cin >> s >> t;
        G[s].push_back(t);
        child[t] = true;
        P[t].push_back(s);
    }
    int root = -1;
    REP(i, 1, N) {
        if (!child[i]){
            root = i;
            break;
        }
    }
    dfs(root);
    REP(i, 1, N) {
        cout << ans[i] << endl;
    }
    

    return 0;
}