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;
vector<long long int> a[100005], deg(100005, 0);
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  while (TESTS--) {
    long long n, m;
    cin >> n >> m;
    for (long long int i = 0; i < m; i++) {
      long long x, y;
      cin >> x >> y;
      a[x].push_back(y);
      a[y].push_back(x);
      deg[x]++;
    }
    priority_queue<long long> pq;
    for (long long int i = 1; i < n + 1; i++)
      if (!deg[i]) pq.push(i);
    vector<long long int> ans(n + 1);
    long long cur = n;
    while (!pq.empty()) {
      long long x = pq.top();
      pq.pop();
      ans[x] = cur;
      cur--;
      for (auto i : a[x]) {
        deg[i]--;
        if (deg[i] == 0) pq.push(i);
      }
    }
    for (long long int i = 1; i < n + 1; i++) cout << ans[i] << " ";
  }
  return 0;
}
