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;
priority_queue<int> pq;
vector<int> adj[100100];
int d[100100];
int ans[100100];
int main() {
  int n, m, u, v;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    cin >> u >> v;
    d[u]++;
    adj[v].push_back(u);
  }
  int big = n;
  for (int i = 1; i <= n; i++)
    if (!d[i]) pq.push(i);
  while (!pq.empty()) {
    u = pq.top();
    pq.pop();
    ans[u] = big--;
    for (int i = 0; i < adj[u].size(); i++) {
      v = adj[u][i];
      d[v]--;
      if (!d[v]) pq.push(v);
    }
  }
  cout << ans[1];
  for (int i = 2; i <= n; i++) cout << ' ' << ans[i];
  cout << '\n';
  return 0;
}
