We are sum for we are manySome 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 $l \le i \lt j \lt k \le r$ and $\operatorname{lcm}(i,j,k) \ge i + j + k$.
Here $\operatorname{lcm}(i, j, k)$ denotes the least common multiple (LCM) of integers $i$, $j$, and $k$.
Input
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 $r$ ($1 \le l \le r \le 2 \cdot 10^5$, $l + 2 \le r$).
Output
For each test case print one integer — the number of suitable triplets.
Note
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:
$(3,4,5)$.
样例
5
1 4
3 5
8 86
68 86
6 86868
3
1
78975
969
109229059713337
题面与样例版权归 Codeforces 及出题人所有;中文翻译由 AI 生成(CI 管线),如有出入请以原文为准。