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 + 5;
int cnt[N], ans[N];
vector<int> lst[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < m; ++i) {
    int u, v;
    cin >> u >> v;
    lst[v].push_back(u);
    ++cnt[u];
  }
  priority_queue<int> pq;
  for (int i = 1; i <= n; ++i) {
    if (cnt[i] == 0) pq.push(i);
  }
  int pt = n;
  while (!pq.empty()) {
    int cur = pq.top();
    pq.pop();
    ans[cur] = pt--;
    for (int i = 0; i < lst[cur].size(); ++i) {
      int nx = lst[cur][i];
      --cnt[nx];
      if (cnt[nx] == 0) pq.push(nx);
    }
  }
  for (int i = 1; i <= n; ++i) {
    cout << ans[i];
    if (i < n) cout << ' ';
  }
  cout << endl;
  return 0;
}
