#include <cmath>
#include <iostream>
using namespace std;
int main(int argc, char **argv) {
  long c, k, m, n, t;
  scanf("%ld", &n);
  t = (long)sqrt(2 * n);
  c = 0;
  for (k = 2; k <= t; k++) {
    if ((2 * n) % k > 0 || (2 * n / k + 1 - k) % 2 > 0)
      continue;
    m = (2 * n / k + 1 - k) / 2;
    c++;
    cout << "%d: %d+...%d\n" << c << m << m + k - 1 << endl;
  }
  cout << "sum is:" << c << endl;
  return 0;
}