human
What is a the Python program you requested given the following:
I found an interesting problem on https://leetcode.com/problems/interval-list-intersections/:
Given two lists of closed intervals, each list of intervals is pairwise disjoint and in sorted order.
Return the intersection of these two interval lists.
(Formally, a closed interval [a, b] (with a <= b) denotes the set of real numbers x with a <= x <= b.  The intersection of two closed intervals is a set of real numbers that is either empty, or can be represented as a closed interval.  For example, the intersection of [1, 3] and [2, 4] is [2, 3].)

 
Example 1:

Input: A = [[0,2],[5,10],[13,23],[24,25]], B = [[1,5],[8,12],[15,24],[25,26]]
Output: [[1,2],[5,5],[8,10],[15,23],[24,24],[25,25]]

 
Note:

0 <= A.length < 1000
0 <= B.length < 1000
0 <= A[i].start, A[i].end, B[i].start, B[i].end < 10^9

I tried it in Python, but could not do it. Can you solve it?
assistant
#Here is the Python program you requested.

class Solution:
    def intervalIntersection(self, A: List[List[int]], B: List[List[int]]) -> List[List[int]]:

        self.k1 = None
        self.k2 = None
        self.res2 = []
        def intersections_append2(k: (int, int), v: int):
            k1 = self.k1
            k2 = self.k2
            if k1 is not None and k2 is not None and k[0] == k1 and k[1] == k2:
                if self.res2:
                    self.res2[-1].append(v)
                    if len(self.res2[-1]) > 2:
                        self.res2[-1] = sorted(set(self.res2[-1]))
                else:
                    self.res2.append([v])
            else:
                self.k1 = k[0]
                self.k2 = k[1]
                self.res2.append([v])
        
        def get_next_items(lft_idx, rgt_idx):
            lft_idx_outer = int(lft_idx / 2)
            lft_idx_inner = lft_idx % 2
            rgt_idx_outer = int(rgt_idx / 2)
            rgt_idx_inner = rgt_idx % 2
            lft = A[lft_idx_outer][lft_idx_inner] if lft_idx_outer < len(A) else None
            rgt = B[rgt_idx_outer][rgt_idx_inner] if rgt_idx_outer < len(B) else None
            lft_full = A[lft_idx_outer] if lft_idx_outer < len(A) else A[lft_idx_outer-1]
            rgt_full = B[rgt_idx_outer] if rgt_idx_outer < len(B) else B[rgt_idx_outer-1]
            return (lft, rgt, lft_full, rgt_full, min(lft_idx_outer, len(A)-1), min(rgt_idx_outer, len(B)-1))
        
        u = (0, 0)
        
        if u[0] >= len(A) or u[0] >= len(B):
            return []
        
        i_a, i_b, next_a2, next_b2, counter_a, counter_b = get_next_items(u[0], u[1])
        
        redo_step = True
        
        while not (i_a is None and i_b is None):
            if (i_a is not None and i_b is None) :
                i, u = i_a, (u[0] + 1, u[1])
            elif (i_a is None and i_b is not None):
                i, u = i_b, (u[0], u[1] + 1)
            elif (i_a < i_b):
                i, u = i_a, (u[0] + 1, u[1])
            elif (i_a > i_b):
                i, u = i_b,...