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: Time Limit Exceeded)

#include <bits/stdc++.h>
#define repr(i,from,to) for(int (i)=(from);(i)<(to);(i)++)
#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)
// const bool debug=true;
const bool debug=false;
#define DEBUG if(debug==true)
#define vprint(x) for(auto a:(x)) cout << x << endl;
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const ll MOD = 1000000007;
template<class T> inline bool chmin(T& a, T b) {
    if (a > b) { a = b; return true;}
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) { a = b; return true;}
    return false;
}
int main(void) {
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n,m;cin>>n>>m;
    set<int> children[n+1];
    vector<bool> is_child(n+1,false);
    vector<int> parent(n+1,0);

    rep(i,n+m-1){
        int s,t;cin>>s>>t;
        s;t;
        is_child[t]=true;
        children[s].insert(t);
    }
    int root;
    repr(i,1,n+1){
        if(!is_child[i]){
            root = i;
        }
    }
    stack<int> que;
    que.push(root);
    vector<bool> check(n,false);
    while(!que.empty()){
        int now = que.top();
        que.pop();
        // if(check[now]){
        //     continue;
        // }
        check[now]=true;
        DEBUG{
            printf("now: %d\n",now);
        }
        for(auto x:children[now]){
            que.push(x);
            parent[x] = now;
        }
    }
    repr(i,1,n+1){
        cout << parent[i] << endl;
    }
    return 0;
}