<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:
        # cnt = Counter(nums)
        n = len(nums)
        # if len(cnt) == 1:
        #   return n**2
        nums.sort()
        s = sum(nums)
        mod = 10**9+7
        res = 0
        for i in range(n):
          a = nums[i]
          if a == 1:
            res += s
            continue
          m = nums[-1]//a
          p = 0
          for j in range(m+1):
            t = a*(j+1)
            idx = bisect_left(nums, t)
            # print(idx,p,idx-p,j)
            res += ((idx-p)*(j))%mod
            res %= mod
            p = idx
          # print(nums[i],res)
        return res %mod
30900