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 N, M;
int ec[114514];
int* es[114514];

void ae(int f, int t) {
	es[f] = realloc(es[f], sizeof(*es[f]) * (ec[f] + 1));
	if (es[f] == NULL) exit(2);
	es[f][ec[f]++] = t;
}

int sansyo[114514];

int next_count[2];
int next_buffer[2][114514];

int answer[114514];

int main(void) {
	int i;
	int phase;
	if (scanf("%d%d", &N, &M) != 2) return 1;
	for (i = 1; i <= N - 1 + M; i++) {
		int A, B;
		if (scanf("%d%d", &A, &B) != 2) return 1;
		ae(A, B);
		sansyo[B]++;
	}
	next_count[0] = 0;
	for (i = 1; i <= N; i++) {
		if (sansyo[i] == 0) next_buffer[0][next_count[0]++] = i;
	}

	phase = 0;
	while (next_count[phase] > 0) {
		next_count[1 - phase] = 0;
		for (i = 0; i < next_count[phase]; i++) {
			int target = next_buffer[phase][i];
			int j;
			for (j = 0; j < ec[target]; j++) {
				if (--sansyo[es[target][j]] == 0) {
					answer[es[target][j]] = target;
					next_buffer[1 - phase][next_count[1 - phase]++] = es[target][j];
				}
			}
		}
		phase = 1 - phase;
	}
	for (i = 1; i <= N; i++) {
		printf("%d\n", answer[i]);
	}
	return 0;
}

/*

一番上(根)はどこからも参照されてない
→ そこからの参照を消せば、次の階層がどこからも参照されなくなるはず
→ 繰り返す

*/