<p>You are given two <strong>0-indexed</strong> integer arrays <code>nums1</code> and <code>nums2</code>, each of size <code>n</code>, and an integer <code>diff</code>. Find the number of <strong>pairs</strong> <code>(i, j)</code> such that:</p>

<ul>
	<li><code>0 &lt;= i &lt; j &lt;= n - 1</code> <strong>and</strong></li>
	<li><code>nums1[i] - nums1[j] &lt;= nums2[i] - nums2[j] + diff</code>.</li>
</ul>

<p>Return<em> the <strong>number of pairs</strong> that satisfy the conditions.</em></p>

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

<pre>
<strong>Input:</strong> nums1 = [3,2,5], nums2 = [2,2,1], diff = 1
<strong>Output:</strong> 3
<strong>Explanation:</strong>
There are 3 pairs that satisfy the conditions:
1. i = 0, j = 1: 3 - 2 &lt;= 2 - 2 + 1. Since i &lt; j and 1 &lt;= 1, this pair satisfies the conditions.
2. i = 0, j = 2: 3 - 5 &lt;= 2 - 1 + 1. Since i &lt; j and -2 &lt;= 2, this pair satisfies the conditions.
3. i = 1, j = 2: 2 - 5 &lt;= 2 - 1 + 1. Since i &lt; j and -3 &lt;= 2, this pair satisfies the conditions.
Therefore, we return 3.
</pre>

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

<pre>
<strong>Input:</strong> nums1 = [3,-1], nums2 = [-2,2], diff = -1
<strong>Output:</strong> 0
<strong>Explanation:</strong>
Since there does not exist any pair that satisfies the conditions, we return 0.
</pre>

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

<ul>
	<li><code>n == nums1.length == nums2.length</code></li>
	<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>-10<sup>4</sup> &lt;= nums1[i], nums2[i] &lt;= 10<sup>4</sup></code></li>
	<li><code>-10<sup>4</sup> &lt;= diff &lt;= 10<sup>4</sup></code></li>
</ul>

class Bit:

    def __init__(self, n: int):
        self.bit = [0 for _ in range(n + 1)]

    def update(self, i: int, val: int) -> None:
        i += 1
        while i < len(self.bit):
            self.bit[i] += val
            i += self._find_RSB(i)

    def prefix_sum(self, i: int) -> int:
        i += 1
        s = 0
        while i > 0:
            s += self.bit[i]
            i -= self._find_RSB(i)
        return s 

    def _find_RSB(self, i: int) -> int:
        return i & -i


class Solution:
    def numberOfPairs(self, nums1: List[int], nums2: List[int], diff: int) -> int:
        res, n = 0, len(nums1)
        shift = 20000
        bit = Bit(100000)
        for i in range(n - 1, -1, -1):
            d = nums2[i] - nums1[i]
            res += bit.prefix_sum(d + diff + shift)
            bit.update(d + shift, 1)
        return res
35900