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 Python (3.4.3) (judge verdict: Wrong Answer)

import sys
from heapq import heappop, heappush
sys.setrecursionlimit(10 ** 9)
input = sys.stdin.readline

N, M = map(int, input().split())
check_lst = [False] * N
lst = [[] for _ in range(N)]
for _ in range(N + M - 1):
    A, B = map(int, input().split())
    A -= 1
    B -= 1
    lst[A].append(B)
    check_lst[B] = True
# print (lst)

ans_lst = [[0, -1] for _ in range(N)] #[距離, 親]を要素とするリスト
#根を求める = どこからも辺が伸びていないところ --> check_lst = Falseの場所
n = check_lst.index(False)
ans_lst[n][1] = -1
# print (n)

que = []
def dfs(i): #iからの経路を探す
    heappush(que, [0, i]) #最初の位置

    while len(que) != 0: #次に行けるところがある限り続ける
        #一番距離が遠くなるところを取り出す
        a = heappop(que) #a[0]: 距離、a[1]: 親
        for i in lst[a[1]]: #i: 有向グラフが繋がっていて次に行けるところ
            if ans_lst[i][0] <= -a[0] + 1: #距離をより長くできるとき
                ans_lst[i][0] = -a[0] + 1 #次に行くべきところの長さを更新
                ans_lst[i][1] = a[1] #次に行くべきところの親は今調べているところ
                heappush(que, [-(a[0] + 1), i])
    return 

dfs(n)
# print (ans_lst)

# print ('以下答え')
for d, i in ans_lst:
    print (i + 1)