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;
vector<vector<int>> vec(100002);
vector<vector<int>> vecRev(100002);
int degree[100002];
int main() {
  int vert, k;
  cin >> vert >> k;
  int a, b;
  for (int i = 0; i < k; i++) {
    cin >> a >> b;
    vec[a].push_back(b);
    vecRev[b].push_back(a);
    degree[a]++;
  }
  priority_queue<int> Q;
  for (int i = 1; i <= vert; i++) {
    if (vec[i].size() == 0) Q.push(i);
  }
  int ans[vert + 2];
  int num = vert;
  while (!Q.empty()) {
    int v = Q.top();
    Q.pop();
    ans[v] = num--;
    for (int i = 0; i < vecRev[v].size(); i++) {
      degree[vecRev[v][i]]--;
      if (degree[vecRev[v][i]] == 0) Q.push(vecRev[v][i]);
    }
  }
  for (int i = 1; i <= vert; i++) cout << ans[i] << " ";
  return 0;
}
