Problem: <image>


A directed acyclic graph (DAG) can be used to represent the ordering of tasks. Tasks are represented by vertices and constraints where one task can begin before another, are represented by edges. For example, in the above example, you can undertake task B after both task A and task B are finished. You can obtain the proper sequence of all the tasks by a topological sort.

Given a DAG $G$, print the order of vertices after the topological sort.

Constraints

* $1 \leq |V| \leq 10,000$
* $0 \leq |E| \leq 100,000$
* There are no parallel edges in $G$
* There are no self loops in $G$

Input

A directed graph $G$ is given in the following format:
$|V|\;|E|$
$s_0 \; t_0$
$s_1 \; t_1$
:
$s_{|E|-1} \; t_{|E|-1}$



$|V|$ is the number of vertices and $|E|$ is the number of edges in the graph. The graph vertices are named with the numbers $0, 1,..., |V|-1$ respectively.

$s_i$ and $t_i$ represent source and target nodes of $i$-th edge (directed).

Output

Print the vertices numbers in order. Print a number in a line.

If there are multiple possible solutions, print any one of them (the solution is judged by a special validator).

Example

Input

6 6
0 1
1 2
3 1
3 4
4 5
5 2


Output

0
3
1
4
5
2

Solution (in PYTHON3):

Solution:
from heapq import heappush, heappop, heapify
from collections import deque, defaultdict, Counter
import itertools
from itertools import permutations, combinations, accumulate, product, combinations_with_replacement
import sys
import bisect
import string
import math
import time

def I(): return int(input())
def S(): return input()
def MI(): return map(int, input().split())
def MS(): return map(str, input().split())
def LI(): return [int(i) for i in input().split()]
def LI_(): return [int(i)-1 for i in input().split()]
def StoI(): return [ord(i)-97 for i in input()]
def ItoS(nn): return chr(nn+97)
def input(): return sys.stdin.readline().rstrip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]

yn = {False: 'No', True: 'Yes'}
YN = {False: 'NO', True: 'YES'}
MOD = 10**9+7
inf = float('inf')
IINF = 10**19
l_alp = string.ascii_lowercase
u_alp = string.ascii_uppercase
ts = time.time()
sys.setrecursionlimit(10**6)
nums = ['1', '2', '3', '4', '5', '6', '7', '8', '9', '10']
show_flg = False
# show_flg = True


def main():
    V, E = MI()
    edges = []

    for i in range(E):
        edges.append(tuple(MI()))

    in_cnt = defaultdict(int)
    outs = defaultdict(list)
    for a, b in edges:
        in_cnt[b] += 1
        outs[a].append(b)

    res = []
    queue = deque([i for i in range(V) if in_cnt[i] == 0])
    while len(queue) != 0:
        v = queue.popleft()
        res.append(v)
        for nxt_v in outs[v]:
            in_cnt[nxt_v] -= 1
            if in_cnt[nxt_v] == 0:
                queue.append(nxt_v)

    for i in range(V):
        print(res[i])

if __name__ == '__main__':
    main()