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;
template <class T>
void splitstr(const string &s, vector<T> &out) {
  istringstream in(s);
  out.clear();
  copy(istream_iterator<T>(in), istream_iterator<T>(), back_inserter(out));
}
template <class T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
static void redirect(int argc, const char **argv) {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  if (argc > 1) {
    static filebuf f;
    f.open(argv[1], ios::in);
    cin.rdbuf(&f);
    if (!cin) {
      cerr << "Failed to open '" << argv[1] << "'" << endl;
      exit(1);
    }
  }
  if (argc > 2) {
    static filebuf f;
    f.open(argv[2], ios::out | ios::trunc);
    cout.rdbuf(&f);
    if (!cout) {
      cerr << "Failed to open '" << argv[2] << "'" << endl;
    }
  }
}
int main(int argc, const char **argv) {
  redirect(argc, argv);
  int N, M;
  cin >> N >> M;
  vector<int> deg(N);
  vector<vector<int>> prior(N);
  for (int i = 0; i < M; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    deg[a]++;
    prior[b].push_back(a);
  }
  priority_queue<int> pq;
  for (int i = 0; i < N; i++)
    if (deg[i] == 0) pq.push(i);
  vector<int> ans(N);
  int nxt = N;
  while (!pq.empty()) {
    int cur = pq.top();
    pq.pop();
    ans[cur] = nxt--;
    for (int v : prior[cur])
      if (--deg[v] == 0) pq.push(v);
  }
  for (int i = 0; i < N; i++) cout << ans[i] << ' ';
  cout << '\n';
  return 0;
}
