#include <bits/stdc++.h>
using namespace std;
const int N = 4005;
const long long INF = 0x3f3f3f3f3f3f3f3f;
template <typename T>
inline void read(T &WOW) {
  T x = 0, flag = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') flag = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  WOW = flag * x;
}
template <typename T>
inline void cmax(T &x, T y) {
  if (x < y) x = y;
}
int n, m, siz[N];
long long a[N], dp[N][N];
int DP(int l, int r) {
  if (l > r) return 0;
  if (l == r) {
    siz[l] = 1;
    dp[l][0] = 0;
    dp[l][1] = a[l] * (m - 1);
    return l;
  }
  int u = l;
  for (int i = l + 1; i <= r; ++i) {
    if (a[i] < a[u]) u = i;
  }
  int v1 = DP(l, u - 1), v2 = DP(u + 1, r);
  siz[u] = min(m, r - l + 1);
  for (int i = 0; i <= siz[u]; ++i) {
    dp[u][i] = -INF;
  }
  for (int i = 0; i <= siz[v1]; ++i) {
    for (int j = 0; j <= siz[v2] && i + j <= siz[u]; ++j) {
      cmax(dp[u][i + j], dp[v1][i] + dp[v2][j] - a[u] * 2 * i * j);
      if (i + j + 1 <= siz[u]) {
        cmax(dp[u][i + j + 1],
             dp[v1][i] + dp[v2][j] - a[u] * (2 * (i * j + i + j) - (m - 1)));
      }
    }
  }
  return u;
}
void solve() {
  read(n);
  read(m);
  for (int i = 1; i <= n; ++i) {
    read(a[i]);
  }
  int rt = DP(1, n);
  printf("%lld\n", dp[rt][m]);
}
int main() {
  solve();
  return 0;
}

### Prompt

Your task is to create a CPP solution to the following problem:

Alice has an integer sequence a of length n and all elements are different. She will choose a subsequence of a of length m, and defines the value of a subsequence a_{b_1},a_{b_2},…,a_{b_m} as $$$∑_{i = 1}^m (m ⋅ a_{b_i}) - ∑_{i = 1}^m ∑_{j = 1}^m f(min(b_i, b_j), max(b_i, b_j)), where f(i, j) denotes \min(a_i, a_{i + 1}, \ldots, a_j)$$$.

Alice wants you to help her to maximize the value of the subsequence she choose.

A sequence s is a subsequence of a sequence t if s can be obtained from t by deletion of several (possibly, zero or all) elements.

Input

The first line contains two integers n and m (1 ≤ m ≤ n ≤ 4000).

The second line contains n distinct integers a_1, a_2, …, a_n (1 ≤ a_i < 2^{31}).

Output

Print the maximal value Alice can get.

Examples

Input


6 4
15 2 18 12 13 4


Output


100


Input


11 5
9 3 7 1 8 12 10 20 15 18 5


Output


176


Input


1 1
114514


Output


0


Input


2 1
666 888


Output


0

Note

In the first example, Alice can choose the subsequence [15, 2, 18, 13], which has the value 4 ⋅ (15 + 2 + 18 + 13) - (15 + 2 + 2 + 2) - (2 + 2 + 2 + 2) - (2 + 2 + 18 + 12) - (2 + 2 + 12 + 13) = 100. In the second example, there are a variety of subsequences with value 176, and one of them is [9, 7, 12, 20, 18].

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 4005;
const long long INF = 0x3f3f3f3f3f3f3f3f;
template <typename T>
inline void read(T &WOW) {
  T x = 0, flag = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') flag = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  WOW = flag * x;
}
template <typename T>
inline void cmax(T &x, T y) {
  if (x < y) x = y;
}
int n, m, siz[N];
long long a[N], dp[N][N];
int DP(int l, int r) {
  if (l > r) return 0;
  if (l == r) {
    siz[l] = 1;
    dp[l][0] = 0;
    dp[l][1] = a[l] * (m - 1);
    return l;
  }
  int u = l;
  for (int i = l + 1; i <= r; ++i) {
    if (a[i] < a[u]) u = i;
  }
  int v1 = DP(l, u - 1), v2 = DP(u + 1, r);
  siz[u] = min(m, r - l + 1);
  for (int i = 0; i <= siz[u]; ++i) {
    dp[u][i] = -INF;
  }
  for (int i = 0; i <= siz[v1]; ++i) {
    for (int j = 0; j <= siz[v2] && i + j <= siz[u]; ++j) {
      cmax(dp[u][i + j], dp[v1][i] + dp[v2][j] - a[u] * 2 * i * j);
      if (i + j + 1 <= siz[u]) {
        cmax(dp[u][i + j + 1],
             dp[v1][i] + dp[v2][j] - a[u] * (2 * (i * j + i + j) - (m - 1)));
      }
    }
  }
  return u;
}
void solve() {
  read(n);
  read(m);
  for (int i = 1; i <= n; ++i) {
    read(a[i]);
  }
  int rt = DP(1, n);
  printf("%lld\n", dp[rt][m]);
}
int main() {
  solve();
  return 0;
}

```