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.

Input

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

Output

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

Examples

Input


3


Output


6


Input


4


Output


18

Note

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 ⋅ 2 ⋅ 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 ⋅ 2 ⋅ 3 + 1 ⋅ 3 ⋅ 4 = 6 + 12 = 18.
#include <bits/stdc++.h>
using namespace std;
template <typename T>
void scan(T& x) {
  x = 0;
  bool neg = 0;
  register T c = getchar();
  if (c == '-') neg = 1, c = getchar();
  while ((c < 48) || (c > 57)) c = getchar();
  for (; c < 48 || c > 57; c = getchar())
    ;
  for (; c > 47 && c < 58; c = getchar()) x = (x << 3) + (x << 1) + (c & 15);
  if (neg) x *= -1;
}
template <typename T>
void print(T n) {
  bool neg = 0;
  if (n < 0) n *= -1, neg = 1;
  char snum[65];
  int i = 0;
  do {
    snum[i++] = n % 10 + '0';
    n /= 10;
  } while (n);
  --i;
  if (neg) putchar('-');
  while (i >= 0) putchar(snum[i--]);
}
string to_string(string s) { return '"' + s + '"'; }
string to_string(const char* s) { return to_string((string)s); }
string to_string(bool b) { return (b ? "true" : "false"); }
string to_string(char ch) { return string("'") + ch + string("'"); }
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <class InputIterator>
string to_string(InputIterator first, InputIterator last) {
  bool start = true;
  string res = "{";
  while (first != last) {
    if (!start) {
      res += ", ";
    }
    start = false;
    res += to_string(*first);
    ++first;
  }
  res += "}";
  return res;
}
template <typename A>
string to_string(A v) {
  bool first = true;
  string res = "{";
  for (const auto& x : v) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(x);
  }
  res += "}";
  return res;
}
void debug_out() { cerr << '\n'; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}
template <typename A, typename B>
istream& operator>>(istream& input, pair<A, B>& x) {
  input >> x.first >> x.second;
  return input;
}
template <typename A>
istream& operator>>(istream& input, vector<A>& x) {
  for (auto& i : x) input >> i;
  return input;
}
mt19937 rng32((int)chrono::steady_clock::now().time_since_epoch().count());
const int mod = 1e9 + 7;
long long gcdExtended(long long a, long long b, long long* x, long long* y);
long long modInverse(long long a, long long m) {
  long long x, y;
  long long g = gcdExtended(a, m, &x, &y);
  if (g != 1)
    cout << "Inverse doesn't exist";
  else {
    long long res = (x % m + m) % m;
    return res;
  }
}
long long gcdExtended(long long a, long long b, long long* x, long long* y) {
  if (a == 0) {
    *x = 0, *y = 1;
    return b;
  }
  long long x1, y1;
  long long gcd = gcdExtended(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return gcd;
}
void solve() {
  long long n;
  cin >> n;
  long long ans = 0;
  for (long long i = 3; i <= n; i++) {
    ans += (i - 1) * (i);
  }
  cout << ans;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
