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>

typedef struct List {
	struct List *next;
	int v;
} list;

int main()
{
	int i, u, w, N, M;
	scanf("%d %d", &N, &M);
	
	int *indeg = (int*)malloc(sizeof(int) * (N + 1));
	list **adj = (list**)malloc(sizeof(list*) * (N + 1)), *d = (list*)malloc(sizeof(list) * (N + M - 1));	
	for (i = 1; i <= N; i++) {
		indeg[i] = 0;
		adj[i] = NULL;
	}
	for (i = 0; i < N + M - 1; i++) {
		scanf("%d %d", &u, &w);
		d[i].v = w;
		d[i].next = adj[u];
		adj[u] = &(d[i]);
		indeg[w]++;
	}
	
	int par[100001] = {}, q[100001], head, tail;
	list *p;
	for (i = 1; i <= N; i++) if (indeg[i] == 0) break;
	q[0] = i;
	par[i] = 0;
	for (head = 0, tail = 1; head < tail; head++) {
		u = q[head];
		for (p = adj[u]; p != NULL; p = p->next) {
			w = p->v;
			indeg[w]--;
			if (indeg[w] == 0) {
				par[w] = u;
				q[tail++] = w;
			}
		}
	}
	
	for (i = 1; i <= N; i++) printf("%d\n", par[i]);
	fflush(stdout);
	return 0;
}