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 long long maxn = 100005;
long long mini[maxn], res[maxn];
set<long long> adj[maxn], radj[maxn];
long long dfs(long long u) {
  if (mini[u] != -1) return mini[u];
  long long res = u;
  for (long long v : adj[u]) {
    res = min(res, dfs(v));
  }
  mini[u] = res;
  return res;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  for (long long i = 0; i < maxn; i++) {
    mini[i] = -1;
  }
  long long n, m;
  cin >> n >> m;
  for (long long i = 0; i < m; i++) {
    long long u, v;
    cin >> u >> v;
    adj[u].insert(v);
    radj[v].insert(u);
  }
  for (long long i = 1; i <= n; i++) dfs(i);
  priority_queue<long long> pq;
  for (long long i = 1; i <= n; i++)
    if (adj[i].empty()) pq.push(i);
  long long cur = n;
  while (!pq.empty()) {
    long long u = pq.top();
    pq.pop();
    res[u] = cur--;
    for (long long v : radj[u]) {
      adj[v].erase(u);
      if (adj[v].empty()) pq.push(v);
    }
  }
  for (long long i = 1; i <= n; i++) cout << res[i] << ' ';
  cout << '\n';
}
