Nebius Welcome Round (Div. 1 + Div. 2)
A
Lame King
Codeforces
800
原题
←B →
翻译中…

You are given a checkerboard of size $201 \times 201$, i. e. it has $201$ rows and $201$ columns. The rows of this checkerboard are numbered from $-100$ to $100$ from bottom to top. The columns of this checkerboard are numbered from $-100$ to $100$ from left to right. The notation $(r, c)$ denotes the cell located in the $r$-th row and the $c$-th column.

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There is a king piece at position $(0, 0)$ and it wants to get to position $(a, b)$ as soon as possible. In this problem our king is lame. Each second, the king makes exactly one of the following five moves.

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  • Skip move. King's position remains unchanged.
  • Go up. If the current position of the king is $(r, c)$ he goes to position $(r + 1, c)$.
  • Go down. Position changes from $(r, c)$ to $(r - 1, c)$.
  • Go right. Position changes from $(r, c)$ to $(r, c + 1)$.
  • Go left. Position changes from $(r, c)$ to $(r, c - 1)$.
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King is

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not allowed

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to make moves that put him outside of the board. The important consequence of the king being lame is that he is

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not allowed

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to make the same move during two consecutive seconds. For example, if the king goes right, the next second he can only skip, go up, down, or left.

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What is the minimum number of seconds the lame king needs to reach position $(a, b)$?

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Input

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The first line of the input contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of test cases. Then follow $t$ lines containing one test case description each.

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Each test case consists of two integers $a$ and $b$ ($-100 \leq a, b \leq 100$) — the position of the cell that the king wants to reach. It is guaranteed that either $a \ne 0$ or $b \ne 0$.

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Output

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Print $t$ integers. The $i$-th of these integers should be equal to the minimum number of seconds the lame king needs to get to the position he wants to reach in the $i$-th test case. The king always starts at position $(0, 0)$.

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Note

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One of the possible solutions for the first example is: go down, go right, go down, go right, go down, go left, go down.

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One of the possible solutions for the second example is to alternate "go right" and "go up" moves $4$ times each.

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One of the possible solutions for the third example is to alternate "go left" and "skip" moves starting with "go left". Thus, "go left" will be used $6$ times, and "skip" will be used $5$ times.

翻译中…

样例

Input #1
5
-4 1
4 4
0 -6
-5 -4
7 -8
Output #1
7
8
11
9
15
InputCopy OutputCopy
题目信息
题号
1804A
时间限制
1 s
内存限制
512 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。