Ivan has number b. He is sorting through the numbers a from 1 to 10^{18}, and for every a writes ([a,    b])/(a) on blackboard. Here [a,    b] stands for least common multiple of a and b. Ivan is very lazy, that's why this task bored him soon. But he is interested in how many different numbers he would write on the board if he would finish the task. Help him to find the quantity of different numbers he would write on the board.

Input

The only line contains one integer — b (1 ≤ b ≤ 10^{10}).

Output

Print one number — answer for the problem.

Examples

Input

1


Output

1

Input

2


Output

2

Note

In the first example [a,    1] = a, therefore ([a,    b])/(a) is always equal to 1.

In the second example [a,    2] can be equal to a or 2 ⋅ a depending on parity of a. ([a,    b])/(a) can be equal to 1 and 2.
#include <bits/stdc++.h>
using namespace std;
map<long long, int> m;
void primeFactors(long long n) {
  while (n % 2 == 0) {
    m[2]++;
    n = n / 2;
  }
  for (int i = 3; i <= sqrt(n); i = i + 2) {
    while (n % i == 0) {
      m[i]++;
      n = n / i;
    }
  }
  if (n > 2) m[n]++;
}
long long gcd(long long a, long long b) {
  if (a % b == 0)
    return b;
  else
    return gcd(b, a % b);
}
int sum(long long a) {
  int sum = 0;
  while (a > 0) {
    sum = sum + (a % 10);
    a = a / 10;
  }
  return sum;
}
int count_digit(long long n) {
  int count = 0;
  while (n > 0) {
    n = n / 10;
    count++;
  }
  return count;
}
int binarySearch(int x, int y, long long z, long long v[]) {
  int low = x;
  int high = y;
  int mid = x + (y - x) / 2;
  while (low <= high) {
    if (v[mid] == z) return mid;
    if (v[mid] < z) return binarySearch(mid + 1, high, z, v);
    if (v[mid] > z) return binarySearch(low, mid - 1, z, v);
  }
  return -1;
}
long long modularExponentiation(long long x, long long n, long long M) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return modularExponentiation((x * x) % M, n / 2, M);
  else
    return (x * modularExponentiation((x * x) % M, (n - 1) / 2, M)) % M;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int tests = 1;
  while (tests--) {
    long long n;
    cin >> n;
    primeFactors(n);
    map<long long, int>::iterator it;
    long long ans = 1;
    for (it = m.begin(); it != m.end(); it++) {
      ans = ans * (it->second + 1);
    }
    cout << ans;
  }
}
