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>
int n, m;
std::vector<int> g[1 + (4 * 100 * 1000)];
std::vector<int> ig[1 + (4 * 100 * 1000)];
int main(void) {
  scanf("%d %d", &n, &m);
  for (int i = m; i--;) {
    int v, u;
    scanf("%d %d", &v, &u);
    g[v].push_back(u);
    ig[u].push_back(v);
  }
  std::priority_queue<int> pq;
  int remaining_out_degree[1 + (4 * 100 * 1000)];
  int label[1 + (4 * 100 * 1000)];
  int current_label = n;
  for (int i = 1; i <= n; i++) {
    remaining_out_degree[i] = g[i].size();
    if (remaining_out_degree[i] == 0) {
      pq.push(i);
    }
  }
  while (!pq.empty()) {
    int v = pq.top();
    pq.pop();
    label[v] = current_label--;
    for (auto u : ig[v]) {
      if (--remaining_out_degree[u] == 0) {
        pq.push(u);
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    printf("%d%c", label[i], (i == n) ? '\n' : ' ');
  }
  return 0;
}
