Given the array `queries`

of positive integers between `1`

and `m`

, you have to process all `queries[i]`

(from `i=0`

to `i=queries.length-1`

) according to the following rules:

- In the beginning, you have the permutation
`P=[1,2,3,...,m]`

. - For the current
`i`

, find the position of`queries[i]`

in the permutation`P`

(**indexing from 0**) and then move this at the beginning of the permutation`P.`

Notice that the position of`queries[i]`

in`P`

is the result for`queries[i]`

.

Return an array containing the result for the given `queries`

.

**Example 1:**

Input:queries = [3,1,2,1], m = 5Output:[2,1,2,1]Explanation:The queries are processed as follow: For i=0: queries[i]=3, P=[1,2,3,4,5], position of 3 in P is2, then we move 3 to the beginning of P resulting in P=[3,1,2,4,5]. For i=1: queries[i]=1, P=[3,1,2,4,5], position of 1 in P is1, then we move 1 to the beginning of P resulting in P=[1,3,2,4,5]. For i=2: queries[i]=2, P=[1,3,2,4,5], position of 2 in P is2, then we move 2 to the beginning of P resulting in P=[2,1,3,4,5]. For i=3: queries[i]=1, P=[2,1,3,4,5], position of 1 in P is1, then we move 1 to the beginning of P resulting in P=[1,2,3,4,5]. Therefore, the array containing the result is [2,1,2,1].

**Example 2:**

Input:queries = [4,1,2,2], m = 4Output:[3,1,2,0]

**Example 3:**

Input:queries = [7,5,5,8,3], m = 8Output:[6,5,0,7,5]

**Constraints:**

`1 <= m <= 10^3`

`1 <= queries.length <= m`

`1 <= queries[i] <= m`

class Solution {
public int[] processQueries(int[] queries, int m) {
}
}