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 mxn = 1e5 + 5;
const int mod = 1e9 + 7;
const int mod1 = 998244353;
int main() {
  ios_base::sync_with_stdio(false);
  int i, n, m, a, b;
  cin >> n >> m;
  int c = n;
  vector<int> v(n), deg(n + 1);
  vector<vector<int> > g(n + 1);
  set<int> s;
  for (i = 0; i < m; i++) {
    cin >> a >> b;
    g[b].push_back(a);
    deg[a]++;
  }
  for (i = 1; i <= n; i++) {
    if (!deg[i]) s.insert(-i);
  }
  while ((int)s.size()) {
    auto x = s.begin();
    v[-1 - *x] = c--;
    s.erase(x);
    for (auto j : g[-*x]) {
      if (--deg[j] == 0) s.insert(-j);
    }
  }
  for (int it = 0; it < (int)v.size(); it++) cout << v[it] << " ";
  cout << endl;
}
