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;
class compare {
 public:
  bool operator()(int x, int y) { return x > y; }
};
int in[100010], ans[100010];
set<int, compare> ms;
set<int>::iterator it;
vector<int> mat[100010];
int main() {
  ios::sync_with_stdio(false);
  int i, j, n, m, a, b, c, d, p;
  cin >> n >> m;
  d = n;
  for (int i = 0; i <= m - 1; i++) {
    cin >> a >> b;
    mat[b].push_back(a);
    in[a]++;
  }
  for (int i = 1; i <= n; i++) {
    if (in[i] == 0) ms.insert(i);
  }
  while (!ms.empty()) {
    it = ms.begin();
    ms.erase(ms.begin());
    p = *it;
    ans[p] = d;
    d--;
    for (int i = 0; i <= (int)mat[p].size() - 1; i++) {
      int x = mat[p][i];
      in[x]--;
      if (in[x] == 0) ms.insert(x);
    }
  }
  for (int i = 1; i <= n; i++) cout << ans[i] << ' ';
  return 0;
}
