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;
int v, u;
vector<int> in[100100];
int out[100100];
set<pair<int, int> > s;
int rj[100100];
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; i++) {
    scanf("%d%d", &v, &u);
    v--;
    u--;
    out[v]++;
    in[u].push_back(v);
  }
  for (int i = 0; i < n; i++) {
    s.insert({out[i], -i});
  }
  for (int k = n; k > 0; k--) {
    pair<int, int> fr = *s.begin();
    s.erase(s.begin());
    assert(fr.first == 0);
    int cv = -fr.second;
    rj[cv] = k;
    for (int i : in[cv]) {
      s.erase({out[i], -i});
      out[i]--;
      s.insert({out[i], -i});
    }
  }
  for (int i = 0; i < n; i++) {
    printf("%d ", rj[i]);
  }
  return 0;
}
