You are given a 2D integer array?items?where?items[i] = [pricei, beautyi]?denotes the?price?and?beauty?of an item respectively.

You are also given a?0-indexed?integer array?queries. For each?queries[j], you want to determine the?maximum beauty?of an item whose?price?is?less than or equal?to?queries[j]. If no such item exists, then the answer to this query is?0.

Return?an array?answer?of the same length as?queries?where?answer[j]?is the answer to the?jth?query.

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Example 1:

Input: items = [[1,2],[3,2],[2,4],[5,6],[3,5]],?

queries = [1,2,3,4,5,6]

Output: [2,4,5,5,6,6]

Explanation:- For queries[0]=1, [1,2] is the only item which has price <= 1. Hence, the answer for this query is 2. - For queries[1]=2, the items which can be considered are [1,2] and [2,4]. ?The maximum beauty among them is 4. - For queries[2]=3 and queries[3]=4, the items which can be considered are [1,2], [3,2], [2,4], and [3,5]. ?The maximum beauty among them is 5. - For queries[4]=5 and queries[5]=6, all items can be considered. ?Hence, the answer for them is the maximum beauty of all items, i.e., 6.

Example 2:

Input: items = [[1,2],[1,2],[1,3],[1,4]],?

queries = [1]

Output: [4]

Explanation: The price of every item is equal to 1, so we choose the item with the maximum beauty 4. Note that multiple items can have the same price and/or beauty. ?

Example 3:

Input: items = [[10,1000]], queries = [5]Output: [0]Explanation:No item has a price less than or equal to 5, so no item can be chosen. Hence, the answer to the query is 0.

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Constraints:

• 1 <= items.length, queries.length <= 105

• items[i].length == 2

• 1 <= pricei, beautyi, queries[j] <= 10

  Easy 題目，直接遍歷一下即可；

Runtime:?1 ms, faster than?91.92%?of?Java?online submissions for?Kids With the Greatest Number of Candies.

Memory Usage:?42.5 MB, less than?32.91%?of?Java?online submissions for?Kids With the Greatest Number of Candies.