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;
inline int GA(long long n, vector<long long>& vec) {
  for (long long i = 0; i < n; ++i) {
    long long x;
    cin >> x;
    vec.push_back(x);
  }
}
long long n, m;
vector<long long> vec[100001];
long long deg[100001];
long long lb[100001];
int main() {
  ios_base::sync_with_stdio(NULL);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  cin >> n >> m;
  for (long long i = 0; i < m; ++i) {
    long long x, y;
    cin >> x >> y;
    x--, y--;
    vec[y].push_back(x);
    deg[x]++;
  }
  priority_queue<long long> p;
  for (long long i = 0; i < n; ++i)
    if (!deg[i]) p.push(i);
  long long num = n;
  while (!p.empty()) {
    long long x = p.top();
    p.pop();
    lb[x] = num--;
    for (auto i : vec[x])
      if (--deg[i] == 0) p.push(i);
  }
  for (long long i = 0; i < n; ++i) cout << lb[i] << " ";
}
