You are living on an infinite plane with the Cartesian coordinate system on it. In one move you can go to any of the four adjacent points (left, right, up, down).

More formally, if you are standing at the point (x,y), you can:

go left, and move to (x?1,y), or go right, and move to (x+1,y)

, or go up, and move to (x,y+1), or go down, and move to (x,y?1).

There are n boxes on this plane. The i-th box has coordinates (xi,yi). It is guaranteed that the boxes are either on the x-axis or the y-axis. That is, either xi=0 or yi=0.

You can collect a box if you and the box are at the same point. Find the minimum number of moves you have to perform to collect all of these boxes if you have to start and finish at the point (0,0)

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你生活在一個(gè)無限平面上，上面有笛卡爾坐標(biāo)系。 您可以一次移動(dòng)到四個(gè)相鄰點(diǎn)（左、右、上、下）中的任何一個(gè)。

更正式地說，如果您站在 (x,y) 點(diǎn)，您可以：

向左移動(dòng)，移動(dòng)到 (x?1,y)，或者向右移動(dòng)，移動(dòng)到 (x+1,y)

，或者向上，移動(dòng)到 (x,y+1)，或者向下，移動(dòng)到 (x,y?1)。

這架飛機(jī)上有n個(gè)盒子。 第 i 個(gè)框的坐標(biāo)為 (xi,yi)。 確保盒子位于 x 軸或 y 軸上。 即，xi=0 或yi=0。

如果你和盒子在同一點(diǎn)，你就可以收集一個(gè)盒子。 如果您必須在點(diǎn) (0,0) 開始和結(jié)束，請(qǐng)求出收集所有這些盒子所需執(zhí)行的最少移動(dòng)次數(shù)。

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其實(shí)就是求x坐標(biāo)軸的最大最小值跟y坐標(biāo)軸的最大最小值即可；