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;

typedef long long ll;

#define p_ary(ary,a,b,i) do { cout << "["; for (int (i) = (a);(i) < (b);++(i)) cout << ary[(i)] << ((b)-1 == (i) ? "" : ", "); cout << "]\n"; } while(0)
#define p_map(map,it) do {cout << "{";for (auto (it) = map.begin();;++(it)) {if ((it) == map.end()) {cout << "}\n";break;}else cout << "" << (it)->first << "=>" << (it)->second << ", ";}}while(0)

int n,m;
vector<vector<int>> edges;
vector<bool> vst;
vector<int> l;

void dfs(vector<vector<int>>& edges, vector<bool>& used, int i) {
    for (int j : edges[i]) if (!used[j]) {
        used[j] = true;
        dfs(edges,used,j);
    }
    l.push_back(i);
}

int main() {
    cin >> n >> m;
    edges.resize(n);
    vector<bool> used(n,false);
    for (int i = 0;i < m+n-1;++i) {
        int a,b;
        cin >> a >> b;
        a--;b--;
        edges[a].push_back(b);
        used[b] = true;
    }
    int par = 0;
    for (int i = 0;i < n;++i) if (!used[i]) par = i;

    vst = vector<bool>(n,false);
    vst[par] = true;
    dfs(edges,vst,par);

    reverse(l.begin(),l.end());
    vector<int> d(n);
    for (int i = 0;i < n;++i) d[l[i]] = i;

    vector<int> pars(n,-1);

    for (int i : l) {
        for (int j : edges[i]) {
            if (pars[j] == -1 || d[pars[j]] < d[i]) {
                pars[j] = i;
            }
        }
    }
    //p_ary(pars,0,n,i);
    for (int i : pars) cout << i+1 << "\n";
    return 0;
}