<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:
        c = Counter(nums)
        m = (10 ** 9) + 7
        nums = sorted(list(set(nums)))
        p = [c[i] for i in nums]
        for i in range(1, len(nums)):
            p[i] += p[i - 1]
        l = len(nums)
        ans = 0
        for i in range(l - 1, -1, -1):
            l = i + 1
            j = 1
            count = 0
            while nums[i] * j <= nums[-1]:
                val = (nums[i] * (j + 1)) - 1
                r = bisect_right(nums, val) - 1
                ln = p[r] - (p[l - 1] if l >= 1 else 0)
                ans = (ans + (c[nums[i]] * ln * j)) % m
                l = r + 1
                j += 1
        for i in c.values():
            ans = (ans + (i ** 2)) % m 
        return ans

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