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 
import sys, heapq
input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__

n, m = map(int, input().split())
g = [[] for _ in range(n + 1)]
indeg = [0] * (n + 1)
for _ in range(m):
    u, v = map(int, input().split())
    g[v].append(u)
    indeg[u] += 1

q = [-u for u in range(1, n + 1) if not indeg[u]]
heapq.heapify(q)
ans = [0] * (n + 1)
cur = n
while q:
    u = -heapq.heappop(q)
    ans[u] = cur
    cur -= 1
    for v in g[u]:
        indeg[v] -= 1
        if not indeg[v]:
            heapq.heappush(q, -v)
print(*ans[1:])