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;
static const int maxn = 100000 + 10;
int n, m, d[maxn], ans[maxn];
vector<int> e[maxn];
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  memset(d, 0, sizeof d);
  for (int i = 0; i < m; i++) {
    int v, u;
    scanf("%d %d", &v, &u);
    e[u].push_back(v);
    d[v]++;
  }
  priority_queue<int> pq;
  for (int i = 1; i <= n; i++)
    if (!d[i]) pq.push(i);
  for (int i = n; i; i--) {
    int u = pq.top();
    ans[u] = i;
    pq.pop();
    for (int j = 0; j < e[u].size(); j++)
      if (!--d[e[u][j]]) pq.push(e[u][j]);
  }
  for (int i = 1; i <= n; i++) printf("%d%c", ans[i], " \n"[i == n]);
  return 0;
}
