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;
int indegree[100022], ans[100020];
vector<int> graph[100020];
int main() {
  int i, j, t;
  scanf("%d%d", &n, &m);
  ;
  for (i = 1; i <= m; i++) {
    int st, en;
    scanf("%d%d", &st, &en);
    ;
    graph[en].push_back(st);
    indegree[st]++;
  }
  set<int> s;
  for (i = 1; i <= n; i++) {
    if (indegree[i] == 0) s.insert(i);
  }
  int counter = 0;
  while (!s.empty()) {
    int ele = *(--s.end());
    ans[ele] = n - counter++;
    s.erase(ele);
    for (i = 0; i < graph[ele].size(); i++) {
      int to = graph[ele][i];
      indegree[to]--;
      if (indegree[to] == 0) s.insert(to);
    }
  }
  for (i = 1; i <= n; i++) {
    printf("%d ", ans[i]);
  }
  cout << endl;
  return 0;
}
