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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, m;
  cin >> n >> m;
  vector<int> outDegree(n, 0);
  vector<vector<int>> adjacencyList(n);
  while (m--) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    adjacencyList[v].emplace_back(u);
    outDegree[u]++;
  }
  priority_queue<int> pq;
  for (int v = 0; v < n; v++) {
    if (!outDegree[v]) pq.emplace(v);
  }
  int currentLabel = n;
  vector<int> label(n);
  while (!pq.empty()) {
    int v = pq.top();
    pq.pop();
    label[v] = currentLabel--;
    for (int u : adjacencyList[v]) {
      if (!--outDegree[u]) pq.emplace(u);
    }
  }
  for (int &lbl : label) cout << lbl << ' ';
  cout << '\n';
}
