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;

int main(){
    int n,m;
    cin >> n >> m;
    vector<int> edge[n+1];
    vector<int> redge[n+1];
    int cnt[n+1];
    fill(cnt,cnt+n+1,0);
    for(int i=0; i<n+m-1; i++){
        int a,b;
        cin >> a >> b;
        edge[a].push_back(b);
        redge[b].push_back(a);
        cnt[b]++;
    }
    queue<int> s;
    for(int i=1; i<n+1; i++){
        if(cnt[i]==0) s.push(i);
    }
    vector<int> vec;
    int rvec[n+1];
    rvec[0] = -1;
    while(!s.empty()){
        int u = s.front();
        s.pop();
        vec.push_back(u);
        rvec[u] = vec.size();
        for(int i=0; i<edge[u].size(); i++){
            cnt[edge[u][i]]--;
            if(cnt[edge[u][i]] == 0) s.push(edge[u][i]);
        }
    }
    int ans[n+1];
    ans[vec[0]] = 0;
    for(int i=1; i<vec.size(); i++){
        int a = vec[i];
        int t = 0;
        for(int j=0; j<redge[a].size(); j++){
            int k = redge[a][j];
            if(rvec[k] > rvec[t]) t = k;
        }
        ans[a] = t;
    }
    for(int i=1; i<n+1; i++){
        cout << ans[i] << endl;
    }
    return 0;
}