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;
const int N = 1e5 + 5;
vector<int> adj[N], order, tdj[N];
int ret[N], deg[N];
int main() {
  int n, i, m, a, b;
  cin >> n >> m;
  for (i = 0; i < m; ++i) {
    scanf("%d%d", &a, &b);
    adj[b].push_back(a);
    ++deg[a];
  }
  priority_queue<int> q;
  for (i = 1; i <= n; ++i)
    if (!deg[i]) q.push(i);
  int pos = 1;
  while (!q.empty()) {
    a = q.top();
    ret[a] = pos++;
    q.pop();
    for (i = 0; i < (int)adj[a].size(); ++i) {
      b = adj[a][i];
      --deg[b];
      if (!deg[b]) {
        q.push(b);
      }
    }
  }
  for (i = 0; i < n; ++i) printf("%d ", n + 1 - ret[i + 1]);
  cout << endl;
}
