human
I found an interesting problem on https://leetcode.com/problems/ugly-number-iii/:
Write a program to find the n-th ugly number.
Ugly numbers are positive integers which are divisible by a or b or c.
 
Example 1:
Input: n = 3, a = 2, b = 3, c = 5
Output: 4
Explanation: The ugly numbers are 2, 3, 4, 5, 6, 8, 9, 10... The 3rd is 4.
Example 2:
Input: n = 4, a = 2, b = 3, c = 4
Output: 6
Explanation: The ugly numbers are 2, 3, 4, 6, 8, 9, 10, 12... The 4th is 6.

Example 3:
Input: n = 5, a = 2, b = 11, c = 13
Output: 10
Explanation: The ugly numbers are 2, 4, 6, 8, 10, 11, 12, 13... The 5th is 10.

Example 4:
Input: n = 1000000000, a = 2, b = 217983653, c = 336916467
Output: 1999999984

 
Constraints:

1 <= n, a, b, c <= 10^9
1 <= a * b * c <= 10^18
It's guaranteed that the result will be in range [1, 2 * 10^9]

I tried it in Python, but could not do it. Can you solve it?
assistant
#Here is Python program to find the n-th ugly number.

class Solution:
    def nthUglyNumber(self, n: int, a: int, b: int, c: int) -> int:
        
        def isEnough(num):
            total = num//a + num//b + num//c - num//ab - num//ac - num//bc + num//abc
            return total >= n
        
        ab = a*b // math.gcd(a,b)
        ac = a*c // math.gcd(a,c)
        bc = b*c // math.gcd(b,c)
        abc = a*bc // math.gcd(a,bc)
        
        left = 1
        right = 10 ** 10
        while left < right:
            mid = left + (right - left) // 2
            if isEnough(mid):
                right = mid
            else:
                left = mid + 1
        return left