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 (Clang 3.8.0) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
#define N_MAX 100001

int main(void){
  int n, m, i, j, p, d, pp, dd, ppn, ddn;
  int ret,a[2][N_MAX], parent[N_MAX];
  ret = scanf("%d %d\n",&n,&m);
  for(i=0;i<n+m;i++){
  	ret = scanf("%d %d\n",&a[0][i],&a[1][i]);
  }

  for(i=1;i<=n;i++){
    p=-1;d=-1;
    for(j=0;j<n+m;j++){
      if(a[1][j] == i){
        if(p==-1){
          p=a[0][j];
        }else{
          d=a[0][j];
        }
      }
      fprintf(stderr, "%d-%d:%d %d \n", i, j, p, d);
    }
    fprintf(stderr, "%d:%d %d \n", i, p, d);
    if(p == -1){
      printf("%d\n",0);
    }else if(d == -1){
      printf("%d\n",p);
    }else{
      pp=p;dd=d;
      while(pp != -1 && dd != -1){
        ppn=-1; ddn=-1;
        for(j=0;j<n+m;j++){
          if(a[1][j] == pp){
            if(ppn == -1)
            ppn=a[0][j];
          }
          if(a[1][j] == dd){
            if(ddn == -1)
            ddn=a[0][j];
          }
        }
        pp=ppn, dd=ddn;
      }
      if(pp != -1){
        printf("%d\n",p);
      }else{
        printf("%d\n",d);
      }
    }
  }
  
  return 0;
}