<p>You are given an integer array <code>coins</code> representing coins of different denominations and an integer <code>k</code>.</p>

<p>You have an infinite number of coins of each denomination. However, you are <strong>not allowed</strong> to combine coins of different denominations.</p>

<p>Return the <code>k<sup>th</sup></code> <strong>smallest</strong> amount that can be made using these coins.</p>

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

<div class="example-block" style="
    border-color: var(--border-tertiary);
    border-left-width: 2px;
    color: var(--text-secondary);
    font-size: .875rem;
    margin-bottom: 1rem;
    margin-top: 1rem;
    overflow: visible;
    padding-left: 1rem;
">
<p><strong>Input:</strong> <span class="example-io" style="
    font-family: Menlo,sans-serif;
    font-size: 0.85rem;
">coins = [3,6,9], k = 3</span></p>

<p><strong>Output:</strong> <span class="example-io" style="
    font-family: Menlo,sans-serif;
    font-size: 0.85rem;
"> 9</span></p>

<p><strong>Explanation:</strong> The given coins can make the following amounts:<br />
Coin 3 produces multiples of 3: 3, 6, 9, 12, 15, etc.<br />
Coin 6 produces multiples of 6: 6, 12, 18, 24, etc.<br />
Coin 9 produces multiples of 9: 9, 18, 27, 36, etc.<br />
All of the coins combined produce: 3, 6, <u><strong>9</strong></u>, 12, 15, etc.</p>
</div>

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

<div class="example-block" style="
    border-color: var(--border-tertiary);
    border-left-width: 2px;
    color: var(--text-secondary);
    font-size: .875rem;
    margin-bottom: 1rem;
    margin-top: 1rem;
    overflow: visible;
    padding-left: 1rem;
">
<p><strong>Input:</strong><span class="example-io" style="
    font-family: Menlo,sans-serif;
    font-size: 0.85rem;
"> coins = [5,2], k = 7</span></p>

<p><strong>Output:</strong><span class="example-io" style="
    font-family: Menlo,sans-serif;
    font-size: 0.85rem;
"> 12 </span></p>

<p><strong>Explanation:</strong> The given coins can make the following amounts:<br />
Coin 5 produces multiples of 5: 5, 10, 15, 20, etc.<br />
Coin 2 produces multiples of 2: 2, 4, 6, 8, 10, 12, etc.<br />
All of the coins combined produce: 2, 4, 5, 6, 8, 10, <u><strong>12</strong></u>, 14, 15, etc.</p>
</div>

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

<ul>
	<li><code>1 &lt;= coins.length &lt;= 15</code></li>
	<li><code>1 &lt;= coins[i] &lt;= 25</code></li>
	<li><code>1 &lt;= k &lt;= 2 * 10<sup>9</sup></code></li>
	<li><code>coins</code> contains pairwise distinct integers.</li>
</ul>

import math
import numpy as np
class Solution:
    def bitsoncount(self,x): 
        return bin(x).count('1') 

    def count(self, a, m, n): 
        odd = 0
        even = 0
        p_list = []
        pow_set_size = 1 << n 
        for counter in range(1, pow_set_size): 
            p=1
            for j in range(n):
                if (counter & (1 << j)):
                    p = p*a[j]//gcd(p,a[j])
            if (self.bitsoncount(counter) & 1): 
                odd += (m // p)
            else: 
                even += (m // p)
            p_list.append((m//p)*p)
        return odd - even , max(p_list)

    def findKthSmallest(self, coins: List[int], k: int) -> int:
        if k == 1:
            return min(coins)

        right_num = min(coins)*k
        left_num = 0
        storage = min(coins)*k
        coins_lis = []
        for i in coins:
            for j in coins_lis:
                if i%j == 0:
                    break
            else:
                if i <= right_num:
                    coins_lis.append(i)
        del coins
        if len(coins_lis) == 1:
            return coins_lis[0]*k

        count = 0
        while True:
            count, ans = self.count(coins_lis, right_num, len(coins_lis))
            if count < k:
                left_num = right_num
                right_num = storage
            elif count > k :
                right_num = left_num + (ans-left_num)//2
            else :
                break
        return ans
1166