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 = 2e6 + 44;
const int X[] = {1, 0, -1, 0, 1, -1, 1, -1};
const int Y[] = {0, 1, 0, -1, 1, 1, -1, -1};
const int mod = 998244353;
const int INF = INT_MAX;
priority_queue<int> s;
vector<int> g[N];
int d[N], ans[N];
int main() {
  int n, m;
  cin >> n >> m;
  int u, v;
  while (m--) {
    cin >> u >> v;
    u--, v--;
    g[v].push_back(u);
    d[u]++;
  }
  for (int i = 0; i < n; i++)
    if (!d[i]) s.push(i);
  int cnt = n;
  while (s.size()) {
    v = s.top();
    s.pop();
    ans[v] = cnt--;
    for (int to : g[v]) {
      d[to]--;
      if (!d[to]) s.push(to);
    }
  }
  for (int i = 0; i < n; i++) cout << ans[i] << " ";
  cout << "\n";
  return 0;
}
