You are given a directed acyclic graph with n vertices and m edges. There are no self-loops or multiple edges between any pair of vertices. Graph can be disconnected.

You should assign labels to all vertices in such a way that:

  * Labels form a valid permutation of length n — an integer sequence such that each integer from 1 to n appears exactly once in it. 
  * If there exists an edge from vertex v to vertex u then labelv should be smaller than labelu. 
  * Permutation should be lexicographically smallest among all suitable. 



Find such sequence of labels to satisfy all the conditions.

Input

The first line contains two integer numbers n, m (2 ≤ n ≤ 105, 1 ≤ m ≤ 105).

Next m lines contain two integer numbers v and u (1 ≤ v, u ≤ n, v ≠ u) — edges of the graph. Edges are directed, graph doesn't contain loops or multiple edges.

Output

Print n numbers — lexicographically smallest correct permutation of labels of vertices.

Examples

Input

3 3
1 2
1 3
3 2


Output

1 3 2 


Input

4 5
3 1
4 1
2 3
3 4
2 4


Output

4 1 2 3 


Input

5 4
3 1
2 1
2 3
4 5


Output

3 1 2 4 5 
from sys import stdin, stdout
from heapq import heappush as hpush, heappop as hpop

n, m = map(int, raw_input().strip().split())
inp = stdin.readlines()
out = [0 for i in xrange(n)]

g = [[] for i in xrange(n + 1)] # the reverse graph
outdeg = [0 for i in xrange(n + 1)] # outdegree

# process input
for line in inp:
    u, v = map(int, line.split())
    outdeg[u] += 1
    g[v].append(u)

# the maxheap
heap = []
for i in xrange(1, n + 1):
    if outdeg[i] == 0: hpush(heap, -i)

# starting with largest label
label = n

# repeat till heap gets empty
while heap:
    u = -hpop(heap)

    out[u - 1] = str(label)
    label -= 1

    for v in g[u]:
        outdeg[v] -= 1

        if outdeg[v] == 0: hpush(heap, -v)

# write output
stdout.write(' '.join(out))