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: Accepted)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
vector<int>v[100005];
set<int>s[100005];
int par[100005];
main()
{
    int n,m,i,j,x,y,ind;
    cin>>n>>m;
    for(i=1;i<=n-1+m;i++)
    {
        cin>>x>>y;
        v[x].push_back(y);
        s[y].insert(x);
    }
    for(i=1;i<=n;i++)
    {
        if(s[i].empty())
        {
            ind=i;break;
        }
    }
    queue<int>q;
    q.push(ind);
    while(!q.empty())
    {
        int top=q.front();q.pop();
        for(int i=0;i<v[top].size();i++)
        {
            int z=v[top][i];
            if(s[z].size()==1)
            {
                  q.push(z);par[z]=top;
            }
            s[z].erase(top);
        }
    }
    for(i=1;i<=n;i++)
    {
        cout<<par[i]<<endl;
    }
}