You are given a 0-indexed integer array `nums` consisting of `3 * n` elements.

You are allowed to remove any subsequence of elements of size exactly `n` from `nums`. The remaining `2 * n` elements will be divided into two equal parts:

- The first `n` elements belonging to the first part and their sum is `sumfirst`.
- The next `n` elements belonging to the second part and their sum is `sumsecond`.

The difference in sums of the two parts is denoted as `sumfirst - sumsecond`.

- For example, if `sumfirst = 3` and `sumsecond = 2`, their difference is `1`.
- Similarly, if `sumfirst = 2` and `sumsecond = 3`, their difference is `-1`.

Return the minimum difference possible between the sums of the two parts after the removal of `n` elements.

Example 1:

    Input: nums = [3,1,2]
    Output: -1
    Explanation: Here, nums has 3 elements, so n = 1.
    Thus we have to remove 1 element from nums and divide the array into two equal parts.
    - If we remove nums[0] = 3, the array will be [1,2]. The difference in sums of the two parts will be 1 - 2 = -1.
    - If we remove nums[1] = 1, the array will be [3,2]. The difference in sums of the two parts will be 3 - 2 = 1.
    - If we remove nums[2] = 2, the array will be [3,1]. The difference in sums of the two parts will be 3 - 1 = 2.
    The minimum difference between sums of the two parts is min(-1,1,2) = -1.

Example 2:

    Input: nums = [7,9,5,8,1,3]
    Output: 1
    Explanation: Here n = 2. So we must remove 2 elements and divide the remaining array into two parts containing two elements each.
    If we remove nums[2] = 5 and nums[3] = 8, the resultant array will be [7,9,1,3]. The difference in sums will be (7+9) - (1+3) = 12.
    To obtain the minimum difference, we should remove nums[1] = 9 and nums[4] = 1. The resultant array becomes [7,5,8,3]. The difference in sums of the two parts is (7+5) - (8+3) = 1.
    It can be shown that it is not possible to obtain a difference smaller than 1.

Constraints:

- `nums.length == 3 * n`
- `1 <= n <= 105`
- `1 <= nums[i] <= 105`