<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>

import math
if hasattr(math,'gcd'):
    gcd = math.gcd
else:
    from fractions import gcd
class Solution:
    def nthUglyNumber(self, n, a, b, c):
        lcm = lambda x,y: x*y//gcd(x,y)
        ab  = lcm(a,b)
        ac  = lcm(a,c)
        bc  = lcm(b,c)
        abc = lcm(ab,c)
        def index(mid):
            Ka,Kb,Kc    = mid//a , mid//b , mid//c
            Kabc        = mid//abc
            Kab,Kac,Kbc = mid//ab, mid//ac, mid//bc
            return Ka + Kb + Kc - Kab - Kac - Kbc + Kabc
        def closest(m):
            Ka,Kb,Kc = m//a, m//b, m//c
            return max(Ka*a,Kb*b,Kc*c)
        lo, hi = min(a,b,c), closest(int(2e9))
        while lo<=hi:
            m   = (lo+hi)//2
            mid = closest( m )
            Kth = index(mid)
            if Kth==n:
                return mid
            if Kth<n:
                lo = m + 1
            else:
                hi = m - 1
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