Given two integer arrays arr1 and arr2, reorder the elements of arr1 so that:

All values that also appear in arr2 come first, and they appear in the same relative order as in arr2. For each value v in arr2, all occurrences of v in arr1 must be placed together.

After placing all elements that are in arr2, append the remaining elements of arr1 that do not appear in arr2, sorted in ascending order.

Return the reordered array.

Note: arr2 contains distinct values, and every value in arr2 appears in arr1.

Constraints:

• $1 \leq$ arr1.length, arr2.length $\leq 1000$

• $0 \leq$ arr1[i], arr2[i] $\leq 1000$

• All elements of arr2 are distinct

• Each arr2[i] is in arr1

Input: arr1 = [5,7,5,8,7,9,1,7,2]-List[int], arr2 = [7,5,2]-List[int]
Output: [7,7,7,5,5,2,1,8,9]-List[int]
Explanation: Values appearing in arr2 come first in arr2 order: all 7s, then all 5s, then all 2s. The remaining values {1,8,9} are appended in ascending order.

Input: arr1 = [10,3,3,6,10,0,6,4,6,2]-List[int], arr2 = [6,10,3]-List[int]
Output: [6,6,6,10,10,3,3,0,2,4]-List[int]
Explanation: Following arr2, all 6s are placed first, then 10s, then 3s. The leftover elements {0,4,2} are sorted to [0,2,4] and appended.

Input: arr1 = [4,4,4,1,2,3,0,5,6,7,8,9]-List[int], arr2 = [4]-List[int]
Output: [4,4,4,0,1,2,3,5,6,7,8,9]-List[int]
Explanation: Only 4 is prioritized, so all 4s come first. All other elements are not in arr2, so they are appended in ascending order.

Input: arr1 = [1000,0,500,500,1000,250,750,250,1]-List[int], arr2 = [250,1000,1]-List[int]
Output: [250,250,1000,1000,1,0,500,500,750]-List[int]
Explanation: Grouping and ordering by arr2 gives all 250s, then all 1000s, then all 1s. The remaining elements {0,500,500,750} are sorted and appended.

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