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 
#include <bits/stdc++.h>
using namespace std;
set<int> a[100001];
int n, m, out[100001], ans[100001];
int main() {
  scanf("%d%d", &n, &m);
  while (m--) {
    int u, v;
    scanf("%d%d", &u, &v);
    a[v].insert(u);
    out[u]++;
  }
  priority_queue<int> q;
  for (int i = 1; i <= n; i++)
    if (!out[i]) q.push(i);
  memset(ans, -1, sizeof ans);
  for (int i = n; i > 0; i--) {
    int u = q.top();
    q.pop();
    for (auto j : a[u])
      if (!--out[j] && ans[j] == -1) q.push(j);
    ans[u] = i;
  }
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  putchar('\n');
  return 0;
}
