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 MaxN = 1e5 + 15;
const int INF = 1e9;
const int MOD = 1e9 + 7;
int n, m;
int cnt[MaxN];
int ans[MaxN];
set<int> s;
vector<int> v[MaxN];
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> m;
  for (int i = 1; i <= m; ++i) {
    int x, y;
    cin >> x >> y;
    v[y].push_back(x);
    ++cnt[x];
  }
  for (int i = 1; i <= n; ++i)
    if (!cnt[i]) s.insert(-i);
  for (int i = n; i >= 1; --i) {
    int x = -(*s.begin());
    ans[x] = i;
    s.erase(s.begin());
    for (int i = 0; i < v[x].size(); ++i) {
      int to = v[x][i];
      --cnt[to];
      if (!cnt[to]) s.insert(-to);
    }
  }
  for (int i = 1; i <= n; ++i) cout << ans[i] << ' ';
  return 0;
}
