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>
#include<iomanip>
#include<numeric>

using namespace std;
using ll = long long;
constexpr int mo = 1e9+7;
constexpr int  mod = mo;
constexpr int inf = 1<<30;
vector<int> E[100050];
vector<int> par(100050,-1);
map<int,int> c;
void bfs(int root){
    queue<int> q;
    q.push(root);
    par[root] = 0;
    while(!q.empty()){
        auto cur = q.front();
        q.pop();
        for(auto e : E[cur]){
            c[e]--;
            if(c[e] == 0){
            par[e] = cur+1;
            q.push(e);
            }
        }
    }
}
int main(){
    int n,m;
    cin>>n>>m;
    int root;
    for(int i=0;i<n+m-1;++i){
        int x,y;
        cin>>x>>y;
        x--;
        y--;
        E[x].push_back(y);
        c[y]++;
    }
    for(int i=0;i<n;++i){
        if(c.count(i) == 0)root = i;
    }
    bfs(root);
    for(int i=0;i<n;++i){
        cout << par[i] << endl;
    }
    return 0;
}