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, h[1000500], cnt, tot, d[1000500], o[1000500];
priority_queue<int, vector<int>, less<int> > q;
inline int rd() {
  int r;
  scanf("%d", &r);
  return r;
}
struct Edge {
  int b, n;
} e[2 * 1000500];
void link(int a, int b) { e[++cnt] = (Edge){b, h[a]}, h[a] = cnt; }
int main() {
  n = rd(), m = rd();
  for (int i = 1; i <= m; i++) {
    int a = rd(), b = rd();
    link(b, a);
    d[a]++;
  }
  for (int i = 1; i <= n; i++)
    if (!d[i]) q.push(i);
  int tot = n;
  while (!q.empty()) {
    int u = q.top();
    q.pop();
    o[u] = tot--;
    for (int i = h[u], v; v = e[i].b, i; i = e[i].n)
      if (--d[v] == 0) q.push(v);
  }
  for (int i = 1; i <= n; i++) printf("%d%c", o[i], i == n ? '\n' : ' ');
  return 0;
}
