You are given a sorted integer array `arr`

containing `1`

and **prime** numbers, where all the integers of `arr`

are unique. You are also given an integer `k`

.

For every `i`

and `j`

where `0 <= i < j < arr.length`

, we consider the fraction `arr[i] / arr[j]`

.

Return *the* `k`

^{th}*smallest fraction considered*. Return your answer as an array of integers of size `2`

, where `answer[0] == arr[i]`

and `answer[1] == arr[j]`

.

**Example 1:**

Input:arr = [1,2,3,5], k = 3Output:[2,5]Explanation:The fractions to be considered in sorted order are: 1/5, 1/3, 2/5, 1/2, 3/5, and 2/3. The third fraction is 2/5.

**Example 2:**

Input:arr = [1,7], k = 1Output:[1,7]

**Constraints:**

`2 <= arr.length <= 1000`

`1 <= arr[i] <= 3 * 10`

^{4}`arr[0] == 1`

`arr[i]`

is a**prime**number for`i > 0`

.- All the numbers of
`arr`

are**unique**and sorted in**strictly increasing**order. `1 <= k <= arr.length * (arr.length - 1) / 2`

class Solution {
public int[] kthSmallestPrimeFraction(int[] arr, int k) {
}
}