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 maxn = 100 * 1000 + 10;
vector<int> g[maxn];
set<pair<int, int> > s;
int cnt[maxn];
int ans[maxn];
int main() {
  int n, m;
  cin >> n >> m;
  int u, v;
  for (int i = 1; i <= n; i++) s.insert({0, -i});
  for (int i = 0; i < m; i++) {
    cin >> u >> v;
    cnt[u]++;
    s.erase({cnt[u] - 1, -u});
    s.insert({cnt[u], -u});
    g[v].push_back(u);
  }
  pair<int, int> node;
  int cv;
  for (int i = 0; i < n; i++) {
    node = (*s.begin());
    cv = -node.second;
    ans[cv] = n - i;
    s.erase(s.begin());
    for (int v : g[cv]) {
      cnt[v]--;
      s.erase({cnt[v] + 1, -v});
      s.insert({cnt[v], -v});
    }
  }
  for (int i = 1; i <= n; i++) {
    cout << ans[i] << ' ';
  }
  cout << endl;
  return 0;
}
