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;
vector<int> grf[200005];
set<pair<int, int> > st;
int outdeg[200005];
int label[200005];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int v, u;
    cin >> v >> u;
    grf[u].push_back(v);
    outdeg[v]++;
  }
  for (int i = 1; i <= n; i++) {
    st.insert(make_pair(outdeg[i], -i));
  }
  int id = n;
  while (!st.empty()) {
    int now = -(*st.begin()).second;
    st.erase(st.begin());
    label[now] = id--;
    for (int i = 0; i < grf[now].size(); i++) {
      int nxt = grf[now][i];
      st.erase(st.find(make_pair(outdeg[nxt], -nxt)));
      outdeg[nxt]--;
      st.insert(make_pair(outdeg[nxt], -nxt));
    }
  }
  for (int i = 1; i <= n; i++) cout << label[i] << " ";
  cout << endl;
  return 0;
}
