Codeforces Round 665 (Div. 2)
B
Ternary Sequence
Codeforces
1100
原题
← AC →
翻译中…

You are given two sequences $a_1, a_2, \dots, a_n$ and $b_1, b_2, \dots, b_n$. Each element of both sequences is either $0$, $1$ or $2$. The number of elements $0$, $1$, $2$ in the sequence $a$ is $x_1$, $y_1$, $z_1$ respectively, and the number of elements $0$, $1$, $2$ in the sequence $b$ is $x_2$, $y_2$, $z_2$ respectively.

翻译中…

You can rearrange the elements in both sequences $a$ and $b$ however you like. After that, let's define a sequence $c$ as follows:

翻译中…

$c_i = \begin{cases} a_i b_i & \mbox{if }a_i \gt b_i \\ 0 & \mbox{if }a_i = b_i \\ -a_i b_i & \mbox{if }a_i \lt b_i \end{cases}$

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You'd like to make $\sum_{i=1}^n c_i$ (the sum of all elements of the sequence $c$) as large as possible. What is the maximum possible sum?

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Input

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The first line contains one integer $t$ ($1 \le t \le 10^4$) — the number of test cases.

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Each test case consists of two lines. The first line of each test case contains three integers $x_1$, $y_1$, $z_1$ ($0 \le x_1, y_1, z_1 \le 10^8$) — the number of $0$-s, $1$-s and $2$-s in the sequence $a$.

翻译中…

The second line of each test case also contains three integers $x_2$, $y_2$, $z_2$ ($0 \le x_2, y_2, z_2 \le 10^8$; $x_1 + y_1 + z_1 = x_2 + y_2 + z_2 \gt 0$) — the number of $0$-s, $1$-s and $2$-s in the sequence $b$.

翻译中…

Output

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For each test case, print the maximum possible sum of the sequence $c$.

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Note

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In the first sample, one of the optimal solutions is:

翻译中…

$a = \{2, 0, 1, 1, 0, 2, 1\}$

翻译中…

$b = \{1, 0, 1, 0, 2, 1, 0\}$

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$c = \{2, 0, 0, 0, 0, 2, 0\}$

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In the second sample, one of the optimal solutions is:

翻译中…

$a = \{0, 2, 0, 0, 0\}$

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$b = \{1, 1, 0, 1, 0\}$

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$c = \{0, 2, 0, 0, 0\}$

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In the third sample, the only possible solution is:

翻译中…

$a = \{2\}$

翻译中…

$b = \{2\}$

翻译中…

$c = \{0\}$

翻译中…

样例

Input #1
3
2 3 2
3 3 1
4 0 1
2 3 0
0 0 1
0 0 1
Output #1
4
2
0
InputCopy OutputCopy
题目信息
题号
1401B
时间限制
2 s
内存限制
256 MB
标签
构造
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。