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;
int n, m, ans[100100], c;
vector<int> o[100100], in[100100];
struct cmp {
  bool operator()(const int& a, const int& b) {
    if (o[a].size() != o[b].size()) return o[a].size() < o[b].size();
    return a > b;
  }
};
set<int, cmp> s;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    o[a - 1].push_back(b - 1);
    in[b - 1].push_back(a - 1);
  }
  for (int i = 0; i < n; i++) s.insert(i);
  c = n;
  while (!s.empty()) {
    int v = *s.begin();
    s.erase(v);
    ans[v] = c;
    c--;
    for (int j = 0; j < in[v].size(); j++) {
      s.erase(in[v][j]);
      for (int k = 0; k < o[in[v][j]].size(); k++)
        if (o[in[v][j]][k] == v) {
          o[in[v][j]].erase(o[in[v][j]].begin() + k);
          break;
        }
      s.insert(in[v][j]);
    }
  }
  for (int i = 0; i < n; i++) printf("%d ", ans[i]);
  printf("\n");
}
