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;
int n, m, v, u, total, k;
int ggraph[100005], ans[100005], llabel[100005], vvertex[100005], a[100005];
priority_queue<int> q;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &v, &u);
    ggraph[v]++;
    total++;
    a[total] = v;
    vvertex[total] = llabel[u];
    llabel[u] = total;
  }
  for (int i = 1; i <= n; i++) {
    if (ggraph[i] == 0) q.push(i);
  }
  k = n;
  while (q.size()) {
    v = q.top();
    q.pop();
    ans[v] = k;
    k--;
    for (int i = llabel[v]; i; i = vvertex[i]) {
      u = a[i];
      if (ggraph[u] > 0) ggraph[u]--;
      if (ggraph[u] == 0) q.push(u);
    }
  }
  for (int i = 1; i <= n; i++) {
    printf("%d", ans[i]);
    if (i < n) printf(" ");
  }
  return 0;
}
