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>
int n, m, a[100001], d[100001];
std::vector<int> e[100001];
std::priority_queue<int> q;
using namespace std;
int main() {
  for (scanf("%d%d", &n, &m); m--;) {
    int u, v;
    scanf("%d%d", &v, &u);
    --u;
    --v;
    e[u].push_back(v);
    ++d[v];
  }
  for (int i = 0; i < n; ++i)
    if (!d[i]) q.push(i);
  for (int i = n; !q.empty();) {
    int c = q.top();
    q.pop();
    a[c] = i--;
    for (auto& u : e[c])
      if (!--d[u]) q.push(u);
  }
  for (int i = 0; i < n; ++i) printf("%d ", a[i]);
  return 0;
}
