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;
const long long N = 2e5 + 10;
void solve() {
  long long n, m;
  cin >> n >> m;
  map<long long, long long> mm;
  map<long long, vector<long long> > in;
  for (long long i = 0; i < m; i++) {
    long long x, y;
    cin >> x >> y;
    in[y].push_back(x);
    mm[x]++;
  }
  priority_queue<long long> can;
  for (long long i = 1; i <= n; i++) {
    if (mm[i] == 0) can.push(i);
  }
  long long ans[n + 1];
  long long curr = n;
  while (!can.empty()) {
    auto t = can.top();
    can.pop();
    ans[t] = curr--;
    for (long long j : in[t]) {
      mm[j]--;
      if (mm[j] == 0) can.push(j);
    }
  }
  for (long long i = 1; i <= n; i++) cout << ans[i] << " ";
}
signed main() {
  long long n = 1;
  while (n--) {
    solve();
  }
}
