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
from collections import defaultdict, deque
from heapq import heappush, heappop
from bisect import bisect_left, bisect_right
import sys
import random
import itertools
import math
sys.setrecursionlimit(10**5)
input = sys.stdin.readline
sqrt = math.sqrt
def LI(): return list(map(int, input().split()))
def LF(): return list(map(float, input().split()))
def LI_(): return list(map(lambda x: int(x)-1, input().split()))


def II(): return int(input())


def IF(): return float(input())


def S(): return input().rstrip()


def LS(): return S().split()


def IR(n): return [II() for _ in range(n)]


def LIR(n): return [LI() for _ in range(n)]


def FR(n): return [IF() for _ in range(n)]


def LFR(n): return [LI() for _ in range(n)]


def LIR_(n): return [LI_() for _ in range(n)]


def SR(n): return [S() for _ in range(n)]


def LSR(n): return [LS() for _ in range(n)]


mod = 1000000007
inf = 1e10

#solve


def solve():
    def dfs(i,j):
        for e in edg[i]:
            if ans[e][1] >= j + 1:
                continue
            dfs(e, j + 1)
            ans[e] = [i, j + 1]
    n, m = LI()
    edg = [[] for i in range(n)]
    rev_edg = [[] for i in range(n)]
    for i in range(n + m - 1):
        a, b = LI_()
        edg[a].append(b)
        rev_edg[b].append(a)
    for i, e in enumerate(rev_edg):
        if len(e):
            continue
        root = i
        break
    ans = [[-1, 0]] * n
    dfs(root, 0)
    print("\n".join(map(str, map(lambda x: x[0] + 1, ans))))

    return


#main
if __name__ == '__main__':
    solve()