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 MAX_N = 100 * 1000 + 10;
int id[MAX_N], d[MAX_N];
vector<int> to[MAX_N];
priority_queue<int> q;
int main() {
  ios_base ::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    to[--v].push_back(--u), d[u]++;
  }
  for (int i = 0; i < n; i++)
    if (!d[i]) q.push(i);
  for (int i = n; i; i--) {
    int u = q.top();
    q.pop(), id[u] = i;
    for (auto v : to[u])
      if (!--d[v]) q.push(v);
  }
  for (int i = 0; i < n; i++) cout << id[i] << ' ';
  return 0;
}
