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 N = 1e5 + 22;
int n, m;
set<int> s;
int d[N], ans[N];
vector<int> G[N];
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  cin >> n >> m;
  for (int i = 0; i < (n); i++) s.insert(i);
  for (int i = 0; i < (m); i++) {
    int x, y;
    cin >> x >> y;
    x--, y--;
    if (d[x] == 0) s.erase(x);
    d[x]++;
    G[y].push_back(x);
  }
  for (int Q = 0; Q < (n); Q++) {
    int x = *s.rbegin();
    ans[x] = n - Q;
    s.erase(x);
    for (int j = 0; j < (G[x].size()); j++) {
      int u = G[x][j];
      if (d[u] == 1) s.insert(u);
      d[u]--;
    }
  }
  for (int i = 0; i < (n); i++) cout << ans[i] << " ";
  return 0;
}
