Preimage size of factorial zeroes function

/*
 * Let f(x) be the number of zeroes at the end of x!. Recall that x! = 1 * 2 * 3 * ... * x and by convention, 0! = 1.
 * For example, f(3) = 0 because 3! = 6 has no zeroes at the end, while f(11) = 2 because 11! = 39916800 has two zeroes at the end.
 * Given an integer k, return the number of non-negative integers x have the property that f(x) = k.
 * Example 1:
 * Input: k = 0
 * Output: 5
 * Explanation: 0!, 1!, 2!, 3!, and 4! end with k = 0 zeroes.
 * Example 2:
 * Input: k = 5
 * Output: 0
 * Explanation: There is no x such that x! ends in k = 5 zeroes.
 * Example 3:
 * Input: k = 3
 * Output: 5
 * Constraints:
 * 0 <= k <= 109
 */
impl Solution {
  pub fn preimage_size_fzf(k: i32) -> i32 {
    fn trailing_zeros(x: i64) -> i64 {
      let mut count = 0i64;
      let mut p = 5i64;
      while p <= x { count += x / p; p *= 5; }
      count
    }
    fn count_le(k: i64) -> i64 {
      if k < 0 { return 0; }
      let mut lo = 0i64;
      let mut hi = 5 * (k + 1);
      while lo < hi {
        let mid = lo + (hi - lo) / 2;
        if trailing_zeros(mid) <= k { lo = mid + 1; } else { hi = mid; }
      }
      lo
    }
    let k = k as i64;
    (count_le(k) - count_le(k - 1)) as i32
  }
}