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 <stdio.h>
#include <stdlib.h>

int main(){
  int n, m;  scanf("%d %d", &n, &m);
  m = n - 1 + m;
  int **graph, to[100005] = {0};
  graph = (int**)malloc(sizeof(int*)*(n+10));
  for(int i=0; i<n+10; i++)
    graph[i] = (int*)malloc(sizeof(int)*(m));
  while(m--){
    int a, b;  scanf("%d %d", &a, &b);
    graph[a][++graph[a][0]] = b;
    to[b]++;
  }

  int queue[100005], front = 0, rear = 0;
  for(int i=1; i<=n; i++)
    if(!to[i])  queue[rear++] = i;

  int topo[100005] = {0};
  while(front != rear){
    int now = queue[front++];
    topo[++topo[0]] = now;
    for(int i=1; i<=graph[now][0]; i++)
      if(--to[graph[now][i]] == 0)
	queue[rear++] = graph[now][i];
  }
  
  int ans[100005] = {0};
  for(int i=1; i<=topo[0]; i++)
    for(int j=1; j<=graph[topo[i]][0]; j++)
      ans[graph[topo[i]][j]] = topo[i];

  for(int i=1; i<=n; i++)
    printf("%d\n", ans[i]);

  return 0;
}