Given an array of positive integers `nums`

, remove the **smallest** subarray (possibly **empty**) such that the **sum** of the remaining elements is divisible by `p`

. It is **not** allowed to remove the whole array.

Return *the length of the smallest subarray that you need to remove, or *`-1`

* if it's impossible*.

A **subarray** is defined as a contiguous block of elements in the array.

**Example 1:**

Input:nums = [3,1,4,2], p = 6Output:1Explanation:The sum of the elements in nums is 10, which is not divisible by 6. We can remove the subarray [4], and the sum of the remaining elements is 6, which is divisible by 6.

**Example 2:**

Input:nums = [6,3,5,2], p = 9Output:2Explanation:We cannot remove a single element to get a sum divisible by 9. The best way is to remove the subarray [5,2], leaving us with [6,3] with sum 9.

**Example 3:**

Input:nums = [1,2,3], p = 3Output:0Explanation:Here the sum is 6. which is already divisible by 3. Thus we do not need to remove anything.

**Example 4:**

Input:nums = [1,2,3], p = 7Output:-1Explanation:There is no way to remove a subarray in order to get a sum divisible by 7.

**Example 5:**

Input:nums = [1000000000,1000000000,1000000000], p = 3Output:0

**Constraints:**

`1 <= nums.length <= 10`

^{5}`1 <= nums[i] <= 10`

^{9}`1 <= p <= 10`

^{9}

class Solution {
public int minSubarray(int[] nums, int p) {
}
}