You are given an integer array nums.

For any positive integer x, define the following sequence:

This sequence will eventually reach the value 1.

The popcount-depth of x is defined as the smallest integer d >= 0 such that pd = 1.

For example, if x = 7 (binary representation "111"). Then, the sequence is: 7 → 3 → 2 → 1, so the popcount-depth of 7 is 3.

You are also given a 2D integer array queries, where each queries[i] is either:

Return an integer array answer, where answer[i] is the number of indices for the ith query of type [1, l, r, k].

 

Example 1:

Input: nums = [2,4], queries = [[1,0,1,1],[2,1,1],[1,0,1,0]]

Output: [2,1]

Explanation:

i queries[i] nums binary(nums) popcount-
depth		 [l, r] k Valid
nums[j]		 updated
nums		 Answer
0 [1,0,1,1] [2,4] [10, 100] [1, 1] [0, 1] 1 [0, 1] — 2
1 [2,1,1] [2,4] [10, 100] [1, 1] — — — [2,1] —
2 [1,0,1,0] [2,1] [10, 1] [1, 0] [0, 1] 0 [1] — 1

Thus, the final answer is [2, 1].

Example 2:

Input: nums = [3,5,6], queries = [[1,0,2,2],[2,1,4],[1,1,2,1],[1,0,1,0]]

Output: [3,1,0]

Explanation:

i queries[i] nums binary(nums) popcount-
depth		 [l, r] k Valid
nums[j]		 updated
nums		 Answer
0 [1,0,2,2] [3, 5, 6] [11, 101, 110] [2, 2, 2] [0, 2] 2 [0, 1, 2] — 3
1 [2,1,4] [3, 5, 6] [11, 101, 110] [2, 2, 2] — — — [3, 4, 6] —
2 [1,1,2,1] [3, 4, 6] [11, 100, 110] [2, 1, 2] [1, 2] 1 [1] — 1
3 [1,0,1,0] [3, 4, 6] [11, 100, 110] [2, 1, 2] [0, 1] 0 [] — 0

Thus, the final answer is [3, 1, 0].

Example 3:

Input: nums = [1,2], queries = [[1,0,1,1],[2,0,3],[1,0,0,1],[1,0,0,2]]

Output: [1,0,1]

Explanation:

i queries[i] nums binary(nums) popcount-
depth		 [l, r] k Valid
nums[j]		 updated
nums		 Answer
0 [1,0,1,1] [1, 2] [1, 10] [0, 1] [0, 1] 1 [1] — 1
1 [2,0,3] [1, 2] [1, 10] [0, 1] — — — [3, 2]  
2 [1,0,0,1] [3, 2] [11, 10] [2, 1] [0, 0] 1 [] — 0
3 [1,0,0,2] [3, 2] [11, 10] [2, 1] [0, 0] 2 [0] — 1

Thus, the final answer is [1, 0, 1].

 

Constraints: