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;
const int N = 200100, M = 100100;
vector<int> gt[N];
int n, m, out[N], foo[N];
set<int> onion;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int u, v, i = 0; i < m; i++) {
    cin >> u >> v;
    gt[v].emplace_back(u);
    out[u]++;
  }
  for (int i = 1; i <= n; i++)
    if (out[i] == 0) onion.emplace(i);
  int rnk = n;
  while (onion.size()) {
    int v = *onion.rbegin();
    onion.erase(v);
    for (auto& u : gt[v]) {
      out[u]--;
      if (out[u] == 0) onion.emplace(u);
    }
    foo[v] = rnk--;
  }
  for (int ey = 1; ey <= n; ey++) cout << foo[ey] << " ";
  cout << '\n';
  return 0;
}
