You are given $n$ points on the plane. The polygon formed from all the $n$ points is strictly convex, that is, the polygon is convex, and there are no three collinear points (i.e. lying in the same straight line). The points are numbered from $1$ to $n$, in clockwise order.

We define the distance between two points $p_1 = (x_1, y_1)$ and $p_2 = (x_2, y_2)$ as their Manhattan distance: $$d(p_1, p_2) = |x_1 - x_2| + |y_1 - y_2|.$$

Furthermore, we define the perimeter of a polygon, as the sum of Manhattan distances between all adjacent pairs of points on it; if the points on the polygon are ordered as $p_1, p_2, \ldots, p_k$ $(k \geq 3)$, then the perimeter of the polygon is $d(p_1, p_2) + d(p_2, p_3) + \ldots + d(p_k, p_1)$.

For some parameter $k$, let's consider all the polygons that can be formed from the given set of points, having any $k$ vertices, such that the polygon is not self-intersecting. For each such polygon, let's consider its perimeter. Over all such perimeters, we define $f(k)$ to be the maximal perimeter.

Please note, when checking whether a polygon is self-intersecting, that the edges of a polygon are still drawn as straight lines. For instance, in the following pictures:

In the middle polygon, the order of points ($p_1, p_3, p_2, p_4$) is not valid, since it is a self-intersecting polygon. The right polygon (whose edges resemble the Manhattan distance) has the same order and is not self-intersecting, but we consider edges as straight lines. The correct way to draw this polygon is ($p_1, p_2, p_3, p_4$), which is the left polygon.

Your task is to compute $f(3), f(4), \ldots, f(n)$. In other words, find the maximum possible perimeter for each possible number of points (i.e. $3$ to $n$).

The first line contains a single integer $n$ ($3 \leq n \leq 3\cdot 10^5$) — the number of points.

Each of the next $n$ lines contains two integers $x_i$ and $y_i$ ($-10^8 \leq x_i, y_i \leq 10^8$) — the coordinates of point $p_i$.

The set of points is guaranteed to be convex, all points are distinct, the points are ordered in clockwise order, and there will be no three collinear points.

For each $i$ ($3\leq i\leq n$), output $f(i)$.