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;
int main() {
  int n, m;
  cin >> n >> m;
  vector<vector<int> > parents(n);
  int deg[n];
  fill(deg, deg + n, 0);
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    parents[b - 1].push_back(a - 1);
    deg[a - 1]++;
  }
  priority_queue<int> q;
  for (int i = 0; i < n; i++) {
    if (deg[i] == 0) q.push(i);
  }
  int rank[n];
  for (int i = n; i >= 1; i--) {
    int x = q.top();
    q.pop();
    for (int j = 0; j < parents[x].size(); j++) {
      deg[parents[x][j]]--;
      if (deg[parents[x][j]] == 0) q.push(parents[x][j]);
    }
    rank[x] = i;
  }
  for (int i = 0; i < n; i++) {
    cout << rank[i] << " ";
  }
  cout << endl;
}
