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 N = 500005;
const long long mod = 1e9 + 7;
vector<long long> adj[N];
vector<bool> vis(N);
vector<long long> in(N), out(N), a(N);
long long n;
void solve() {
  long long x = 0, y = 0, c = 0, ans;
  long long m, k;
  cin >> n >> m;
  ans = n;
  for (long long i = 0; i < m; ++i) {
    cin >> x >> y;
    adj[y].push_back(x);
    out[x]++;
    in[y]++;
  }
  priority_queue<long long> nodes;
  for (long long i = 1; i <= n; ++i) {
    if (out[i] == 0) nodes.push(i);
  }
  while (!nodes.empty()) {
    x = nodes.top();
    nodes.pop();
    a[x] = ans--;
    for (auto i : adj[x]) {
      out[i]--;
      if (out[i] == 0) nodes.push(i);
    }
  }
  for (long long i = 1; i <= n; ++i) {
    cout << a[i] << " ";
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
  return 0;
}
