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;
priority_queue<int> que;
int to[100005], pre[100005], lst[100005], deg[100005], tot, ans[100005];
inline void add_edge(int u, int v) {
  to[tot] = v;
  pre[tot] = lst[u];
  deg[v]++;
  lst[u] = tot++;
}
int main() {
  memset(lst, -1, sizeof(lst));
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    add_edge(--v, --u);
  }
  for (int i = 0; i < n; i++) {
    if (!deg[i]) que.push(i);
  }
  int cur = n;
  while (!que.empty()) {
    int u = que.top();
    que.pop();
    ans[u] = cur--;
    for (int i = lst[u]; ~i; i = pre[i]) {
      if (!--deg[to[i]]) que.push(to[i]);
    }
  }
  for (int i = 0; i < n; i++) printf("%d%c", ans[i], " \n"[i + 1 == n]);
  return 0;
}
