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;
bool cmp(const pair<int, int>& a, const pair<int, int>& b) {
  return a.first < b.first || (a.first == b.first && a.second > b.second);
}
vector<int> adj[112345];
int main() {
  int n, m;
  cin >> n >> m;
  int degree[n + 1], ans[n + 1];
  for (int i = 1; i <= n; i++) {
    degree[i] = 0;
    adj[i].clear();
  }
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    adj[v].push_back(u);
    degree[u]++;
  }
  set<pair<int, int>,
      bool (*)(const pair<int, int>& a, const pair<int, int>& b)>
      s(&cmp);
  s.clear();
  for (int i = 1; i <= n; i++) {
    s.insert(make_pair(degree[i], i));
  }
  for (int i = n; i > 0; i--) {
    pair<int, int> curr = *s.begin();
    s.erase(s.begin());
    ans[curr.second] = i;
    for (int i = 0; i < adj[curr.second].size(); i++) {
      int v = adj[curr.second][i];
      pair<int, int> val;
      val = *s.find(make_pair(degree[v], v));
      s.erase(make_pair(degree[v], v));
      val.first--;
      degree[val.second]--;
      s.insert(val);
    }
  }
  for (int i = 1; i <= n; i++) {
    if (i > 1) cout << " ";
    cout << ans[i];
  }
  cout << endl;
}
