<p>Given an integer array <code>nums</code>, return the sum of <code>floor(nums[i] / nums[j])</code> for all pairs of indices <code>0 &lt;= i, j &lt; nums.length</code> in the array. Since the answer may be too large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>The <code>floor()</code> function returns the integer part of the division.</p>

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

<pre>
<strong>Input:</strong> nums = [2,5,9]
<strong>Output:</strong> 10
<strong>Explanation:</strong>
floor(2 / 5) = floor(2 / 9) = floor(5 / 9) = 0
floor(2 / 2) = floor(5 / 5) = floor(9 / 9) = 1
floor(5 / 2) = 2
floor(9 / 2) = 4
floor(9 / 5) = 1
We calculate the floor of the division for every pair of indices in the array then sum them up.
</pre>

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

<pre>
<strong>Input:</strong> nums = [7,7,7,7,7,7,7]
<strong>Output:</strong> 49
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution:
    def sumOfFlooredPairs(self, nums: List[int]) -> int:
      n,ans = 2*max(nums) + 11,0
      p,arr = [0 for _ in range(n)],[]
      for i in nums:
        p[i] += 1 
        if p[i] == 1 : arr.append(i)
      arr.sort()
      for i in range(1,n): p[i] += p[i-1] 
      for x in arr : 
        i, j, curr = x, x, 0 
        for i in range(x,n,x):
          ans += (p[x] - p[x-1]) * (i//x -1) * (p[i-1]-p[j-1]) 
          ans, j = ans%1000000007, i
      
      return ans
32800