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> q;
int ne[100010], he[100010], e[100010], cnt, in[100010], ans[100010];
void add(int u, int v) {
  ne[++cnt] = he[u];
  he[u] = cnt;
  e[cnt] = v;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    add(v, u);
    in[u]++;
  }
  for (int i = 1; i <= n; i++)
    if (in[i] == 0) q.push(i);
  int tt = n;
  while (!q.empty()) {
    int tmp = q.top();
    q.pop();
    ans[tmp] = tt--;
    for (int i = he[tmp]; i; i = ne[i])
      if ((--in[e[i]]) == 0) q.push(e[i]);
  }
  for (int i = 1; i < n; i++) printf("%d ", ans[i]);
  printf("%d\n", ans[n]);
  return 0;
}
