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, …, p_k (k ≥ 3), then the perimeter of the polygon is d(p_1, p_2) + d(p_2, p_3) + … + 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:

<image>

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), …, f(n). In other words, find the maximum possible perimeter for each possible number of points (i.e. 3 to n).

Input

The first line contains a single integer n (3 ≤ n ≤ 3⋅ 10^5) — the number of points. 

Each of the next n lines contains two integers x_i and y_i (-10^8 ≤ x_i, y_i ≤ 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.

Output

For each i (3≤ i≤ n), output f(i).

Examples

Input

4
2 4
4 3
3 0
1 3


Output

12 14 

Input

3
0 0
0 2
2 0


Output

8 

Note

In the first example, for f(3), we consider four possible polygons: 

  * (p_1, p_2, p_3), with perimeter 12. 
  * (p_1, p_2, p_4), with perimeter 8. 
  * (p_1, p_3, p_4), with perimeter 12. 
  * (p_2, p_3, p_4), with perimeter 12. 



For f(4), there is only one option, taking all the given points. Its perimeter 14.

In the second example, there is only one possible polygon. Its perimeter is 8.
#include <bits/stdc++.h>
using namespace std;
struct point {
  long long x, y;
} p[300005], l, r, u, d;
int n;
int cmp1(point &a, point &b) { return a.x < b.x; }
int cmp2(point &a, point &b) { return a.x > b.x; }
int cmp3(point &a, point &b) { return a.y > b.y; }
int cmp4(point &a, point &b) { return a.y < b.y; }
int equ(point a, point b) { return a.x == b.x && a.y == b.y; }
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> p[i].x >> p[i].y;
  sort(p + 1, p + 1 + n, cmp1);
  l = p[1];
  sort(p + 1, p + 1 + n, cmp2);
  r = p[1];
  sort(p + 1, p + 1 + n, cmp3);
  u = p[1];
  sort(p + 1, p + 1 + n, cmp4);
  d = p[1];
  long long Max = 0;
  for (int i = 1; i <= n; i++) {
    if (equ(p[i], l) || equ(p[i], r)) continue;
    point k1 = p[i], k2 = l, k3 = r;
    long long up = max(k1.y, max(k2.y, k3.y));
    long long down = min(k1.y, min(k2.y, k3.y));
    long long ri = max(k1.x, max(k2.x, k3.x));
    long long le = min(k1.x, min(k2.x, k3.x));
    Max = max(Max, 2 * (ri - le) + 2 * (up - down));
  }
  for (int i = 1; i <= n; i++) {
    if (equ(p[i], u) || equ(p[i], d)) continue;
    point k1 = p[i], k2 = u, k3 = d;
    long long up = max(k1.y, max(k2.y, k3.y));
    long long down = min(k1.y, min(k2.y, k3.y));
    long long ri = max(k1.x, max(k2.x, k3.x));
    long long le = min(k1.x, min(k2.x, k3.x));
    Max = max(Max, 2 * (ri - le) + 2 * (up - down));
  }
  for (int i = 1; i <= n; i++) {
    if (equ(p[i], l) || equ(p[i], d)) continue;
    point k1 = p[i], k2 = l, k3 = d;
    long long up = max(k1.y, max(k2.y, k3.y));
    long long down = min(k1.y, min(k2.y, k3.y));
    long long ri = max(k1.x, max(k2.x, k3.x));
    long long le = min(k1.x, min(k2.x, k3.x));
    Max = max(Max, 2 * (ri - le) + 2 * (up - down));
  }
  for (int i = 1; i <= n; i++) {
    if (equ(p[i], l) || equ(p[i], u)) continue;
    point k1 = p[i], k2 = l, k3 = u;
    long long up = max(k1.y, max(k2.y, k3.y));
    long long down = min(k1.y, min(k2.y, k3.y));
    long long ri = max(k1.x, max(k2.x, k3.x));
    long long le = min(k1.x, min(k2.x, k3.x));
    Max = max(Max, 2 * (ri - le) + 2 * (up - down));
  }
  for (int i = 1; i <= n; i++) {
    if (equ(p[i], r) || equ(p[i], d)) continue;
    point k1 = p[i], k2 = r, k3 = d;
    long long up = max(k1.y, max(k2.y, k3.y));
    long long down = min(k1.y, min(k2.y, k3.y));
    long long ri = max(k1.x, max(k2.x, k3.x));
    long long le = min(k1.x, min(k2.x, k3.x));
    Max = max(Max, 2 * (ri - le) + 2 * (up - down));
  }
  for (int i = 1; i <= n; i++) {
    if (equ(p[i], r) || equ(p[i], u)) continue;
    point k1 = p[i], k2 = r, k3 = u;
    long long up = max(k1.y, max(k2.y, k3.y));
    long long down = min(k1.y, min(k2.y, k3.y));
    long long ri = max(k1.x, max(k2.x, k3.x));
    long long le = min(k1.x, min(k2.x, k3.x));
    Max = max(Max, 2 * (ri - le) + 2 * (up - down));
  }
  cout << Max << " ";
  for (int i = 4; i <= n; i++) cout << 2 * (u.y - d.y) + 2 * (r.x - l.x) << " ";
}
