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: Time Limit Exceeded)

import sys

sys.setrecursionlimit(100000000)


# p:parent
def bfs(g: {}, v: int, p: int, depth: int, dd: {}):
    d_v = dd.setdefault(v, [])
    d_v.append((p, depth))

    if v in g.keys():
        for next_v in g[v]:
            # next_v が親だったらスルー
            if next_v == p:
                continue
            # 再帰的に探索
            bfs(g, next_v, v, depth + 1, dd)


def main():
    n, m = map(int, input().split())

    children = set()
    g = {}
    parent_g = {}
    for i in range(n - 1 + m):
        a, b = map(int, input().split())
        dic_a = g.setdefault(a - 1, [])
        dic_a.append(b - 1)
        children.add(b - 1)
        dic_b = parent_g.setdefault(b - 1, [])
        dic_b.append(a - 1)

    root = -1
    for i in range(n):
        if i not in children:
            root = i
            break
    parent = -1
    depth = 0
    dd = {}
    bfs(g, root, parent, depth, dd)

    for i in range(n):
        v = dd[i]
        if len(v) == 1:
            print(v[0][0]+1)
            continue

        v.sort(key=lambda x: x[1], reverse=True)
        print(v[0][0]+1)



main()