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]
class Solution:
    def nthUglyNumber(self, n: int, a: int, b: int, c: int) -> int:
        ab = a*b//gcd(a, b)
        bc = b*c//gcd(b, c)
        ca = c*a//gcd(c, a)
        abc = ab*c//gcd(ab, c)
        
        def fn(k):
            return k//a + k//b + k//c - k//ab - k//bc - k//ca + k//abc

        lo, hi = 0, 2_000_000_000
        while lo < hi:
            mid = (lo + hi)//2
            if fn(mid) >= n: hi = mid
            else: lo = mid + 1
        return lo
class Solution:
    def nthUglyNumber(self, n: int, a: int, b: int, c: int) -> int:
        ab = a*b//gcd(a, b)
        bc = b*c//gcd(b, c)
        ca = c*a//gcd(c, a)
        abc = ab*c//gcd(ab, c)
        
        def fn(k):
            return k//a + k//b + k//c - k//ab - k//bc - k//ca + k//abc

        lo, hi = 0, 2_000_000_000
        while lo < hi:
            mid = (lo + hi)//2
            if fn(mid) >= n: hi = mid
            else: lo = mid + 1
        return lo