Problem: Mr. Santa asks all the great programmers of the world to solve a trivial problem. He gives them an integer m and asks for the number of positive integers n, such that the factorial of n ends with exactly m zeroes. Are you among those great programmers who can solve this problem?

Input

The only line of input contains an integer m (1 ≤ m ≤ 100 000) — the required number of trailing zeroes in factorial.

Output

First print k — the number of values of n such that the factorial of n ends with m zeroes. Then print these k integers in increasing order.

Examples

Input

1


Output

5
5 6 7 8 9 

Input

5


Output

0

Note

The factorial of n is equal to the product of all integers from 1 to n inclusive, that is n! = 1·2·3·...·n.

In the first sample, 5! = 120, 6! = 720, 7! = 5040, 8! = 40320 and 9! = 362880.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
template <typename T>
bool mmax(T &m, const T q) {
  if (m < q) {
    m = q;
    return true;
  } else
    return false;
}
template <typename T>
bool mmin(T &m, const T q) {
  if (m > q) {
    m = q;
    return true;
  } else
    return false;
}
void __print(long long first) { cerr << first; }
void __print(long first) { cerr << first; }
void __print(unsigned first) { cerr << first; }
void __print(unsigned long first) { cerr << first; }
void __print(unsigned long long first) { cerr << first; }
void __print(float first) { cerr << first; }
void __print(double first) { cerr << first; }
void __print(long double first) { cerr << first; }
void __print(char first) { cerr << '\'' << first << '\''; }
void __print(const char *first) { cerr << '\"' << first << '\"'; }
void __print(const string &first) { cerr << '\"' << first << '\"'; }
void __print(bool first) { cerr << (first ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &first) {
  cerr << '{';
  __print(first.first);
  cerr << ',';
  __print(first.second);
  cerr << '}';
}
template <typename T>
void __print(const T &first) {
  long long f = 0;
  cerr << '{';
  for (auto &i : first) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
long long dx[] = {0, 0, 1, -1};
long long dy[] = {1, -1, 0, 0};
long long go(long long first) {
  long long ans = 0;
  long long p = 5;
  while (1) {
    ans += first / p;
    if (first / p == 0) break;
    p *= 5;
  }
  return ans;
}
vector<long long> v;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long m;
  cin >> m;
  long long l = 0, h = (long long)1e18, lf = h, rf = -1;
  while (l <= h) {
    long long mid = (l + h) >> 1;
    long long d = go(mid);
    if (d == m) {
      lf = mid;
      h = mid - 1;
    } else if (d > m) {
      h = mid - 1;
    } else
      l = mid + 1;
  }
  l = 0, h = (long long)1e18;
  while (l <= h) {
    long long mid = (l + h) >> 1;
    long long d = go(mid);
    if (d == m) {
      rf = mid;
      l = mid + 1;
    } else if (d > m) {
      h = mid - 1;
    } else
      l = mid + 1;
  }
  if (rf == -1) {
    cout << 0 << "\n";
    return 0;
  }
  cout << rf - lf + 1 << "\n";
  for (long long i = lf; i <= rf; i++) cout << i << " ";
  return 0;
}