Problem: Ksusha is a vigorous mathematician. She is keen on absolutely incredible mathematical riddles. 

Today Ksusha came across a convex polygon of non-zero area. She is now wondering: if she chooses a pair of distinct points uniformly among all integer points (points with integer coordinates) inside or on the border of the polygon and then draws a square with two opposite vertices lying in the chosen points, what will the expectation of this square's area be?

A pair of distinct points is chosen uniformly among all pairs of distinct points, located inside or on the border of the polygon. Pairs of points p, q (p ≠ q) and q, p are considered the same.

Help Ksusha! Count the required expectation.

Input

The first line contains integer n (3 ≤ n ≤ 105) — the number of vertices of Ksusha's convex polygon. Next n lines contain the coordinates of the polygon vertices in clockwise or counterclockwise order. The i-th line contains integers xi, yi (|xi|, |yi| ≤ 106) — the coordinates of the vertex that goes i-th in that order.

Output

Print a single real number — the required expected area. 

The answer will be considered correct if its absolute and relative error doesn't exceed 10 - 6.

Examples

Input

3
0 0
5 5
5 0


Output

4.6666666667


Input

4
-1 3
4 5
6 2
3 -5


Output

8.1583333333


Input

3
17 136
859 937
16 641


Output

66811.3704155169

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct ww {
  double x, s, t, sumy, sumyy;
} a[2000010];
int i, j, k, n, m;
int x[2000010], y[2000010];
double l[2000010], r[2000010], s, t, an;
inline void work(int x, int y, int xx, int yy) {
  if (x > xx) swap(x, xx), swap(y, yy);
  if (x == xx) {
    l[x] = min(l[x], 0. + y);
    r[x] = max(r[x], 0. + yy);
    return;
  }
  int i;
  for (i = x; i <= xx; i++) {
    double k = 1. * (yy - y) * (i - x) / (xx - x) + y;
    l[i] = min(l[i], k);
    r[i] = max(r[i], k);
  }
}
inline void Do() {
  int i;
  for (i = 0; i <= 2000010 - 1; i++) r[i] = 0, l[i] = 2000010;
  for (i = 1; i <= n; i++) work(x[i], y[i], x[i + 1], y[i + 1]);
  m = 0;
  for (i = 0; i <= 2000010 - 1; i++)
    if (l[i] < 2000010) {
      int A = int(l[i] - (1e-11)) + 1, B = int(r[i] + (1e-11));
      m++;
      a[m].x = i;
      a[m].s = B - A + 1;
      a[m].t = a[m].x * a[m].s;
    }
  s = t = 0;
  for (i = 1; i <= m; i++) s += a[i].s, t += a[i].t;
  for (i = 1; i <= m; i++)
    an += a[i].t * a[i].x * (s - a[i].s) - (a[i].t * (t - a[i].t));
}
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++)
    scanf("%d%d", &x[i], &y[i]), x[i] += 2000010 / 2, y[i] += 2000010 / 2;
  x[n + 1] = x[1], y[n + 1] = y[1];
  Do();
  for (i = 1; i <= n + 1; i++) swap(x[i], y[i]);
  Do();
  an /= s * (s - 1);
  printf("%.10lf\n", an);
  return 0;
}