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;
void solve() {
  long long int n, m;
  cin >> n >> m;
  vector<long long int> adj[n + 1];
  vector<long long int> indegree(n + 1);
  for (long long int i = 0; i < m; i++) {
    long long int u, v;
    cin >> u >> v;
    adj[v].push_back(u);
    indegree[u]++;
  }
  priority_queue<long long int> q;
  for (long long int i = 1; i <= n; i++) {
    if (indegree[i] == 0) {
      q.push(i);
    }
  }
  long long int put = n;
  vector<long long int> ans(n + 1);
  while (!q.empty()) {
    long long int u = q.top();
    q.pop();
    ans[u] = put;
    put--;
    for (long long int v : adj[u]) {
      indegree[v]--;
      if (indegree[v] == 0) {
        q.push(v);
      }
    }
  }
  for (long long int i = 1; i <= n; i++) {
    cout << ans[i] << " ";
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}
