Given a **(0-indexed)** integer array `nums`

and two integers `low`

and `high`

, return *the number of nice pairs*.

A **nice pair** is a pair `(i, j)`

where `0 <= i < j < nums.length`

and `low <= (nums[i] XOR nums[j]) <= high`

.

**Example 1:**

Input:nums = [1,4,2,7], low = 2, high = 6Output:6Explanation:All nice pairs (i, j) are as follows: - (0, 1): nums[0] XOR nums[1] = 5 - (0, 2): nums[0] XOR nums[2] = 3 - (0, 3): nums[0] XOR nums[3] = 6 - (1, 2): nums[1] XOR nums[2] = 6 - (1, 3): nums[1] XOR nums[3] = 3 - (2, 3): nums[2] XOR nums[3] = 5

**Example 2:**

Input:nums = [9,8,4,2,1], low = 5, high = 14Output:8Explanation:All nice pairs (i, j) are as follows: - (0, 2): nums[0] XOR nums[2] = 13 - (0, 3): nums[0] XOR nums[3] = 11 - (0, 4): nums[0] XOR nums[4] = 8 - (1, 2): nums[1] XOR nums[2] = 12 - (1, 3): nums[1] XOR nums[3] = 10 - (1, 4): nums[1] XOR nums[4] = 9 - (2, 3): nums[2] XOR nums[3] = 6 - (2, 4): nums[2] XOR nums[4] = 5

**Constraints:**

`1 <= nums.length <= 2 * 10`

^{4}`1 <= nums[i] <= 2 * 10`

^{4}`1 <= low <= high <= 2 * 10`

^{4}

class Solution {
public int countPairs(int[] nums, int low, int high) {
}
}