You are given an integer N. You have to find how many numbers (X) exist for the given N such that 1≤x≤N and the number of divisors of X is a power of 2 .
Input format
Output format
Print a single integer denoting the number of valid 1≤x≤N for each query as space-separated integers.
Constraints
For N=3
Number of divisors of 1=1 (can be represented as 20)
Number of divisors of 2=2 (can be represented as 21)
Number of divisors of 3=2 (can be represented as 21)
Therefore, the total count of valid x(1≤x≤3) is 3.