Codeforces Round 841 (Div. 2) and Divide by Zero 2022
B
Kill Demodogs
Codeforces
1100
原题
← AC →
翻译中…

Demodogs from the Upside-down have attacked Hawkins again. El wants to reach Mike and also kill as many Demodogs in the way as possible.

翻译中…

Hawkins can be represented as an $n \times n$ grid. The number of Demodogs in a cell at the $i$-th row and the $j$-th column is $i \cdot j$. El is at position $(1, 1)$ of the grid, and she has to reach $(n, n)$ where she can find Mike.

翻译中…

The only directions she can move are the right (from $(i, j)$ to $(i, j + 1)$) and the down (from $(i, j)$ to $(i + 1, j)$). She can't go out of the grid, as there are doors to the Upside-down at the boundaries.

翻译中…

Calculate the maximum possible number of Demodogs $\mathrm{ans}$ she can kill on the way, considering that she kills all Demodogs in cells she visits (including starting and finishing cells).

翻译中…

Print $2022 \cdot \mathrm{ans}$ modulo $10^9 + 7$. Modulo $10^9 + 7$ because the result can be too large and multiplied by $2022$ because we are never gonna see it again!

翻译中…

(Note, you firstly multiply by $2022$ and only after that take the remainder.)

翻译中…

Input

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Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \leq t \leq 10^4$). Description of the test cases follows.

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The first line of each test case contains one integer $n$ ($2 \leq n \leq 10^9$) — the size of the grid.

翻译中…

Output

翻译中…

For each test case, print a single integer — the maximum number of Demodogs that can be killed multiplied by $2022$, modulo $10^9 + 7$.

翻译中…

Note

翻译中…

In the first test case, for any path chosen by her the number of Demodogs to be killed would be $7$, so the answer would be $2022 \cdot 7 = 14154$.

翻译中…

样例

Input #1
4
2
3
50
1000000000
Output #1
14154
44484
171010650
999589541
InputCopy OutputCopy
题目信息
题号
1731B
时间限制
1 s
内存限制
256 MB
标签
贪心
数学

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。