Given an integer array nums, you can perform the following operation any number of times.

In one operation, select a positive integer x such that at least one element of nums is nonzero, and for every index i where nums[i] > 0, replace nums[i] with nums[i] - x.

Return the minimum number of operations required to make every element in nums equal to $0$.

Note: In each operation, x must be a positive value that does not make any currently positive element become negative.

Constraints:

• $1 \leq$ nums.length $\leq 100$

• $0 \leq$ nums[i] $\leq 100$

Input: nums = [1, 5, 0, 3, 5]-List[int]
Output: 3-int
Explanation: Only positive elements are reduced each time, so the minimum number of operations equals the number of distinct positive values, {1, 3, 5}, which is 3. One optimal sequence is choosing x = 1 ([0, 4, 0, 2, 4]), then x = 2 ([0, 2, 0, 0, 2]), then x = 2 ([0, 0, 0, 0, 0]).

Input: nums = [0, 0, 0]-List[int]
Output: 0-int
Explanation: All elements are already zero, so no operations are needed.

Input: nums = [2, 2, 2]-List[int]
Output: 1-int
Explanation: There is only one distinct positive value, {2}, so choosing x = 2 gives [0, 0, 0] in one operation.

Input: nums = [4, 1, 4, 2]-List[int]
Output: 3-int
Explanation: The distinct positive values are {1, 2, 4}, so 3 operations are needed. One optimal sequence is choosing x = 1 ([3, 0, 3, 1]), then x = 1 ([2, 0, 2, 0]), then x = 2 ([0, 0, 0, 0]).

Input: nums = [0, 7, 7, 1, 0, 3]-List[int]
Output: 3-int
Explanation: The distinct positive values are {1, 3, 7}, so 3 operations are needed. A valid sequence is x = 1 ([0, 6, 6, 0, 0, 2]), then x = 2 ([0, 4, 4, 0, 0, 0]), then x = 4 ([0, 0, 0, 0, 0, 0]).

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