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:
#
# 　　  ⋀_⋀　 
#　　  (･ω･)  
# .／ Ｕ ∽ Ｕ＼
#  │＊　合　＊│
#  │＊　格　＊│ 
#  │＊　祈　＊│ 
#  │＊　願　＊│ 
#  │＊　　　＊│ 
#      ￣
#
import sys
sys.setrecursionlimit(10**6)
input=sys.stdin.readline
from math import floor,ceil,sqrt,factorial,log #log2ないｙｐ
from heapq import heappop, heappush, heappushpop
from collections import Counter,defaultdict,deque
from itertools import accumulate,permutations,combinations,product,combinations_with_replacement
from bisect import bisect_left,bisect_right
from copy import deepcopy
inf=float('inf')
mod = 10**9+7
def pprint(*A): 
    for a in A:     print(*a,sep='\n')
def INT_(n): return int(n)-1
def MI(): return map(int,input().split())
def MF(): return map(float, input().split())
def MI_(): return map(INT_,input().split())
def LI(): return list(MI())
def LI_(): return [int(x) - 1 for x in input().split()]
def LF(): return list(MF())
def LIN(n:int): return [I() for _ in range(n)]
def LLIN(n: int): return [LI() for _ in range(n)]
def LLIN_(n: int): return [LI_() for _ in range(n)]
def LLI(): return [list(map(int, l.split() )) for l in input()]
def I(): return int(input())
def F(): return float(input())
def ST(): return input().replace('\n', '')
def main():
    N,M = MI()
    edge = [[]for _ in range(N)]
    indegree = [0]*N #入次数
    for _ in range(M):
        a,b = MI()
        edge[a].append(b)
        indegree[b] += 1
    #print(indegree)
    #入次数が0の点からあれしていく
    queue = deque() 
    for i,x in enumerate(indegree):
        if not x:
            queue.append(i)

    pop = queue.popleft
    topological_index = []
    while queue:
        v = pop()
        topological_index.append(v)

        for to in edge[v]:
            indegree[to] -= 1
            if not indegree[to]:
                queue.append(to)
    if tuple(1 for i in indegree if i!=0):
        print("loop")

    print(*topological_index,sep="\n")



if __name__ == '__main__':
    main()