Codeforces Round 813 (Div. 2)
E2
LCM Sum (hard version)
Codeforces
2500
原题
← E1F →
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We are sum for we are manySome Number

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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.

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You are given two positive integers $l$ and $r$.

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Count the number of distinct triplets of integers $(i, j, k)$ such that $l \le i \lt j \lt k \le r$ and $\operatorname{lcm}(i,j,k) \ge i + j + k$.

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Here $\operatorname{lcm}(i, j, k)$ denotes the least common multiple (LCM) of integers $i$, $j$, and $k$.

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Input

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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.

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The only line for each test case contains two integers $l$ and $r$ ($1 \le l \le r \le 2 \cdot 10^5$, $l + 2 \le r$).

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Output

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

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Note

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In the first test case, there are $3$ suitable triplets:

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  • $(1,2,3)$,
  • $(1,3,4)$,
  • $(2,3,4)$.
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In the second test case, there is $1$ suitable triplet:

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  • $(3,4,5)$.
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样例

Input #1
5
1 4
3 5
8 86
68 86
6 86868
Output #1
3
1
78975
969
109229059713337
InputCopy OutputCopy
题目信息
题号
1712E2
时间限制
3.5 s
内存限制
512 MB
标签
暴力
数据结构
数学
数论
双指针

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。