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;
vector<int> g[100005];
int deg[100005];
int ans[100005];
int main() {
  int n, m;
  int u, v;
  cin >> n >> m;
  for (int i = 0; i < m; ++i) {
    cin >> v >> u;
    ++deg[v];
    g[u].push_back(v);
  }
  priority_queue<int> q;
  for (int i = 1; i < n + 1; ++i) {
    if (deg[i] == 0) {
      q.push(i);
    }
  }
  for (int i = n; i >= 1; --i) {
    u = q.top();
    q.pop();
    ans[u] = i;
    for (int j = 0; j < g[u].size(); ++j) {
      v = g[u][j];
      --deg[v];
      if (deg[v] == 0) {
        q.push(v);
      }
    }
  }
  for (int i = 1; i < n + 1; ++i) cout << ans[i] << ' ';
  cout << endl;
  return 0;
}
