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 REP(i,n) for(int i=0,_n=(int)(n);i<_n;++i)
#define ALL(v) (v).begin(),(v).end()
#define CLR(t,v) memset(t,(v),sizeof(t))
template<class T1,class T2>ostream& operator<<(ostream& os,const pair<T1,T2>&a){return os<<"("<<a.first<<","<<a.second<< ")";}
template<class T>void pv(T a,T b){for(T i=a;i!=b;++i)cout<<(*i)<<" ";cout<<endl;}
template<class T>void chmin(T&a,const T&b){if(a>b)a=b;}
template<class T>void chmax(T&a,const T&b){if(a<b)a=b;}


int nextInt() { int x; scanf("%d", &x); return x;}

const int MAX_N = 112345;
int indeg[MAX_N];
vector<int> g[MAX_N];
int ans[MAX_N];

int main2() {
  CLR(indeg, 0);
  REP(i, MAX_N) g[i].clear();
  CLR(ans, -1);
  int N = nextInt();
  int M = nextInt();
  REP(i, N+M-1) {
    int a = nextInt() - 1;
    int b = nextInt() - 1;
    g[a].push_back(b);
    ++indeg[b];
  }
  deque< pair<int,int> > q; // v, prev
  REP(i, N)if (indeg[i] == 0) { q.push_back( {i, -1} ); }
  while (!q.empty()) {
    auto cur = q.front(); q.pop_front();
    // cout << cur << endl;
    ans[ cur.first ] = cur.second + 1;
    for (auto u : g[cur.first]) {
      --indeg[u];
      if (indeg[u] == 0) {
        q.push_back({ u, cur.first });
      }
    }
  }
  REP(i, N) printf("%d\n", ans[i]);
  return 0;
}

int main() {
  for (;!cin.eof();cin>>ws)
    main2();
  return 0;
}