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++ (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include <bits/stdc++.h>
#define REP(i, s, n) for (int i = s; i < (int)(n); i++)
#define ALL(a) a.begin(), a.end()
#define MOD 1000000007
using namespace std;
using ll = long long;

int N, M;
vector<vector<int>> G;
vector<pair<int, int>> D;

void dfs(int v) {
    for (auto g : G[v]) {
        if (D[g].first < D[v].first + 1) {
            D[g] = {D[v].first + 1, v};
            dfs(g);
        }
    }
}

void dijkstra(int v) {
    D[v] = {0, -1};

    using pi = pair<int, int>;
    priority_queue<pi> pq;
    pq.push({0, v});

    while (!pq.empty()) {
        auto q = pq.top(); pq.pop();
        // cout << "# v : " << q.second << ", dist : " << q.first << " , D : " << D[q.second].first << endl;
        if (q.first < D[q.second].first) continue;
        for (auto g : G[q.second]) {
            if (D[g].first < q.first + 1) {
                // cout << "--> next : " << g << endl;
                D[g] = {q.first + 1, q.second};
                pq.push({D[g].first, g});
            }
        }
    }
}

int main() {
    cin >> N >> M;
    G.resize(N); vector<int> cnt(N, 0);
    REP(i, 0, N - 1 + M) {
        int u, v; cin >> u >> v;
        G[--u].push_back(--v);
        cnt[v]++;
    }

    D.resize(N);
    REP(i, 0, N) if (cnt[i] == 0) dijkstra(i);

    REP(i, 0, N) cout << D[i].second + 1 << endl;
    return 0;
}