Minimum Triangulation

给出一个逆时针顶点标号为 $1\sim n(n\leq 500)$ 的正多边形，一个三角形划分的权值是三角形三个顶点的编号乘积之和。 输出一个最小的权值。

You are given a regular polygon with $ n $ vertices labeled from $ 1 $ to $ n $ in counter-clockwise order. The triangulation of a given polygon is a set of triangles such that each vertex of each triangle is a vertex of the initial polygon, there is no pair of triangles such that their intersection has non-zero area, and the total area of all triangles is equal to the area of the given polygon. The weight of a triangulation is the sum of weigths of triangles it consists of, where the weight of a triagle is denoted as the product of labels of its vertices. Calculate the minimum weight among all triangulations of the polygon.

The first line contains single integer $ n $ ( $ 3 \le n \le 500 $ ) — the number of vertices in the regular polygon.

Print one integer — the minimum weight among all triangulations of the given polygon.

3

6

4

18

According to Wiki: polygon triangulation is the decomposition of a polygonal area (simple polygon) $ P $ into a set of triangles, i. e., finding a set of triangles with pairwise non-intersecting interiors whose union is $ P $ . In the first example the polygon is a triangle, so we don't need to cut it further, so the answer is $ 1 \cdot 2 \cdot 3 = 6 $ . In the second example the polygon is a rectangle, so it should be divided into two triangles. It's optimal to cut it using diagonal $ 1-3 $ so answer is $ 1 \cdot 2 \cdot 3 + 1 \cdot 3 \cdot 4 = 6 + 12 = 18 $ .