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 PyPy3 (2.4.0) (judge verdict: Time Limit Exceeded)

#!/usr/bin/env python3
import sys
import heapq as hq
sys.setrecursionlimit(100000)
N, M = map(int, input().split())
g = [[] for _ in range(N)]
rg = [[] for _ in range(N)]
for _ in range(N - 1 + M):
    A, B = map(lambda x: int(x) - 1, input().split())
    g[A].append(B)
    rg[B].append(A)


def dfs(s):
    global ans
    for t in rg[s]:
        return dfs(t)
    else:
        ans[s] = 0
        return s


def bfs(s, dep):
    global ans
    global dist
    h = [[-dep, s]]
    while h:
        d, u = hq.heappop(h)
        if dist[u] > -d:
            continue
        for v in g[u]:
            if dist[v] <= dist[u] + 1:
                ans[v] = u + 1
                dist[v] = dist[u] + 1
                hq.heappush(h, [-(-d + 1), v])


ans = [None] * N
dist = [0] * N
s = dfs(0)
bfs(s, 0)

for x in ans:
    print(x)