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>
std::priority_queue<int> Q;
int check[100010];
std::vector<int> V[100010], V2[100010];
int score[100010];
int main() {
  int a, b;
  scanf("%d%d", &a, &b);
  for (int i = 1; i <= b; i++) {
    int c, d;
    scanf("%d%d", &c, &d);
    V[c].push_back(d);
    V2[d].push_back(c);
  }
  for (int i = 1; i <= a; i++) score[i] = V[i].size();
  for (int i = 1; i <= a; i++)
    if (score[i] == 0) Q.push(i);
  int C = a;
  while (!Q.empty()) {
    int k = Q.top();
    Q.pop();
    check[k] = C--;
    for (int i = 0; i < V2[k].size(); i++) {
      score[V2[k][i]]--;
      if (score[V2[k][i]] == 0) Q.push(V2[k][i]);
    }
  }
  for (int i = 1; i <= a; i++) printf("%d ", check[i]);
}
