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 = (int)1e6 + 100;
const int mod = (int)1e9 + 7;
const int maxlog = (int)20;
const int P = mod;
int n, m, res, ans[maxn];
vector<int> g[maxn], gr[maxn];
set<pair<int, int> > second;
int main() {
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    int u, v;
    cin >> u >> v;
    g[v].push_back(u);
    gr[u].push_back(v);
  }
  res = n;
  for (int i = 1; i <= n; i++) {
    second.insert(make_pair(gr[i].size(), -i));
  }
  while (second.size()) {
    int v = second.begin()->second;
    v = -v;
    second.erase(second.begin());
    if (ans[v]) continue;
    ans[v] = res--;
    for (int to : g[v]) {
      if (!ans[to]) {
        gr[to].erase(gr[to].begin());
        second.insert(make_pair(gr[to].size(), -to));
      }
    }
  }
  for (int i = 1; i <= n; i++) cout << ans[i] << " ";
}
