CF1712E2 LCM Sum (hard version)

We are sum for we are many Some Number This version of the problem differs from the previous one only in the constraint on $ t $ . You can make hacks only if both versions of the problem are solved. You are given two positive integers $ l $ and $ r $ . Count the number of distinct triplets of integers $ (i, j, k) $ such that $ l \le i < j < k \le r $ and $ \operatorname{lcm}(i,j,k) \ge i + j + k $ . Here $ \operatorname{lcm}(i, j, k) $ denotes the [least common multiple (LCM)](https://en.wikipedia.org/wiki/Least_common_multiple) of integers $ i $ , $ j $ , and $ k $ .

Each test contains multiple test cases. The first line contains the number of test cases $ t $ ( $ \bf{1 \le t \le 10^5} $ ). Description of the test cases follows. The only line for each test case contains two integers $ l $ and $ r $ ( $ 1 \le l \le r \le 2 \cdot 10^5 $ , $ l + 2 \le r $ ).

For each test case print one integer — the number of suitable triplets.

In the first test case, there are $ 3 $ suitable triplets: - $ (1,2,3) $ , - $ (1,3,4) $ , - $ (2,3,4) $ . In the second test case, there is $ 1 $ suitable triplet: - $ (3,4,5) $ .