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 INF = 1e9 + 23;
const int MOD = 1e9 + 7;
vector<int> g[100000 + 23];
int label[100000 + 23];
int d[100000 + 23];
set<int> s;
int main() {
  int n, m, x, y;
  int c;
  cin >> n >> m;
  c = n;
  for (int i = 1; i <= m; i++) {
    cin >> x >> y;
    g[y].push_back(x);
    d[x]++;
  }
  for (int i = 1; i <= n; i++) {
    if (d[i] == 0) {
      s.insert(i);
    }
  }
  while (s.size() != 0) {
    int u = *(s.rbegin());
    s.erase(u);
    for (int i = 0; i < g[u].size(); i++) {
      int v = g[u][i];
      d[v]--;
      if (d[v] == 0) {
        s.insert(v);
      }
    }
    label[u] = c;
    c--;
  }
  for (int i = 1; i <= n; i++) {
    cout << label[i] << " ";
  }
  return 0;
}
