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 II() {
  int n;
  scanf("%d", &n);
  return n;
}
void IIb(int n) { printf("%d ", n); }
constexpr int MAXN = 100000;
std::vector<int> in_adj[MAXN];
int out_degree[MAXN];
int ans[MAXN];
int main() {
  int N = II(), M = II(), f = N;
  while (M--) {
    int u = II() - 1, v = II() - 1;
    in_adj[v].push_back(u);
    ++out_degree[u];
  }
  std::priority_queue<int> pq;
  for (int u = 0; u < N; ++u)
    if (!out_degree[u]) pq.push(u);
  while (!pq.empty()) {
    int u = pq.top();
    pq.pop();
    ans[u] = f--;
    for (int v : in_adj[u])
      if (!--out_degree[v]) pq.push(v);
  }
  for (int u = 0; u < N; ++u) IIb(ans[u]);
  putchar(10);
}
