<p>An <strong>ugly number</strong> is a positive integer that is divisible by <code>a</code>, <code>b</code>, or <code>c</code>.</p>

<p>Given four integers <code>n</code>, <code>a</code>, <code>b</code>, and <code>c</code>, return the <code>n<sup>th</sup></code> <strong>ugly number</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 3, a = 2, b = 3, c = 5
<strong>Output:</strong> 4
<strong>Explanation:</strong> The ugly numbers are 2, 3, 4, 5, 6, 8, 9, 10... The 3<sup>rd</sup> is 4.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 4, a = 2, b = 3, c = 4
<strong>Output:</strong> 6
<strong>Explanation:</strong> The ugly numbers are 2, 3, 4, 6, 8, 9, 10, 12... The 4<sup>th</sup> is 6.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 5, a = 2, b = 11, c = 13
<strong>Output:</strong> 10
<strong>Explanation:</strong> The ugly numbers are 2, 4, 6, 8, 10, 11, 12, 13... The 5<sup>th</sup> is 10.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n, a, b, c &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= a * b * c &lt;= 10<sup>18</sup></code></li>
	<li>It is guaranteed that the result will be in range <code>[1, 2 * 10<sup>9</sup>]</code>.</li>
</ul>

class Solution:
    def nthUglyNumber(self, n: int, a: int, b: int, c: int) -> int:

        def gcd(x, y):
            while y:
                x, y = y, x % y
            return x
        
        def lcm(x, y):
            return x * y // gcd(x, y)
        
        def count(mid):

            lcm_ab = lcm(a, b)
            lcm_ac = lcm(a, c)
            lcm_bc = lcm(b, c)
            lcm_abc = lcm(lcm_ab, c)

            return mid // a + mid // b + mid // c - mid // lcm_ab - mid // lcm_ac - mid // lcm_bc + mid // lcm_abc >= n

        left, right, index = 0, 2 * 10 ** 9, 0

        while left <= right:
            mid = left + right >> 1
            if count(mid):
                index = mid
                right = mid - 1
            else:
                left = mid + 1
        return index
        
        
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