BEST PYTHON SOLUTION WITH 1062MS OF RUNTIME !
\n# Code\npython3 []\nfrom itertools import accumulate\nfrom bisect import bisect_right\nfrom typing import List\n\nclass Solution:\n    def gcdValues(self, num
\n# Code\n```python3 []\nfrom itertools import accumulate\nfrom bisect import bisect_right\nfrom typing import List\n\nclass Solution:\n    def gcdValues(self, nums: List[int], queries: List[int]) -> List[int]:\n        maxv = max(nums)\n        co = [0] * (maxv + 1)\n        lcount = [0] * (maxv + 1)\n        \n        # Count the occurrence of each number in nums\n        for v in nums:\n            co[v] += 1\n        \n        # Calculate lcount for each divisor\n        for i in range(1, maxv + 1):\n            x = 0\n            for mul in range(i, maxv + 1, i):\n                x += co[mul]\n            lcount[i] = x * (x - 1) // 2\n        \n        # Adjust lcount to subtract over-counted GCDs\n        for i in range(maxv, 0, -1):\n            for mul in range(2 * i, maxv + 1, i):\n                lcount[i] -= lcount[mul]\n        \n        # Accumulate counts for answering queries efficiently\n        acc = list(accumulate(lcount))\n        \n        # Answer each query by finding the appropriate GCD count using binary search\n        return [bisect_right(acc, q) for q in queries]\n\n```
['Python3']