You are given an integer array nums, an integer array queries, and an integer x.
For each queries[i], you need to find the index of the queries[i]th occurrence of x in the nums array. If there are fewer than queries[i] occurrences of x, the answer should be -1 for that query.
Return an integer array answer containing the answers to all queries.
\n
Example 1:
\n\nInput: nums = [1,3,1,7], queries = [1,3,2,4], x = 1
\n\nOutput: [0,-1,2,-1]
\n\nExplanation:
\n\nnums, so the answer is -1.nums, so the answer is -1.Example 2:
\n\nInput: nums = [1,2,3], queries = [10], x = 5
\n\nOutput: [-1]
\n\nExplanation:
\n\nnums, so the answer is -1.\n
Constraints:
\n\n1 <= nums.length, queries.length <= 1051 <= queries[i] <= 1051 <= nums[i], x <= 104给你一个整数数组 nums ，一个整数数组 queries 和一个整数 x 。
对于每个查询 queries[i] ，你需要找到 nums 中第 queries[i] 个 x 的位置，并返回它的下标。如果数组中 x 的出现次数少于 queries[i] ，该查询的答案为 -1 。
请你返回一个整数数组 answer ，包含所有查询的答案。
\n\n
示例 1：
\n\n输入：nums = [1,3,1,7], queries = [1,3,2,4], x = 1
\n\n输出：[0,-1,2,-1]
\n\n解释：
\n\nnums 中只有两个 1 ，所以答案为 -1 。nums 中只有两个 1 ，所以答案为 -1 。示例 2：
\n\n输入：nums = [1,2,3], queries = [10], x = 5
\n\n输出：[-1]
\n\n解释：
\n\nnums 中没有 5 ，所以答案为 -1 。\n\n
提示：
\n\n1 <= nums.length, queries.length <= 1051 <= queries[i] <= 1051 <= nums[i], x <= 104nums and save all the occurrences of each element in the separate arrays."
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