#include <bits/stdc++.h>
#pragma GCC optimize(2, 3, "Ofast")
using namespace std;
template <typename T1, typename T2>
void ckmin(T1 &a, T2 b) {
  if (a > b) a = b;
}
template <typename T1, typename T2>
void ckmax(T1 &a, T2 b) {
  if (a < b) a = b;
}
int read() {
  int x = 0, f = 0;
  char ch = getchar();
  while (!isdigit(ch)) f |= ch == '-', ch = getchar();
  while (isdigit(ch)) x = 10 * x + ch - '0', ch = getchar();
  return f ? -x : x;
}
template <typename T>
void print(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) print(x / 10);
  putchar(x % 10 + '0');
}
template <typename T>
void print(T x, char let) {
  print(x), putchar(let);
}
const int N = 4005;
long long dp[N][N];
int sz[N], a[N], n, m, tot;
void solve(int &u, int l, int r) {
  if (l > r) {
    u = 0;
    return;
  }
  u = ++tot, sz[u] = r - l + 1;
  dp[u][0] = 0;
  dp[u][1] = 1ll * (m - 1) * a[l];
  int pos = l, ls = 0, rs = 0;
  for (int i = l; i <= r; i++)
    if (a[i] < a[pos]) pos = i;
  solve(ls, l, pos - 1), solve(rs, pos + 1, r);
  if (ls && rs) {
    for (int i = 1; i <= sz[ls]; i++) {
      for (int j = 1; j <= sz[rs]; j++) {
        ckmax(dp[u][i + j], dp[ls][i] + dp[rs][j] - 2ll * a[pos] * i * j);
        ckmax(dp[u][i + j + 1], dp[ls][i] + dp[rs][j] + 1ll * (m - 1) * a[pos] -
                                    2ll * a[pos] * (1ll * i * j + i + j));
      }
    }
  }
  if (ls) {
    for (int i = 0; i <= sz[ls]; i++) {
      ckmax(dp[u][i], dp[ls][i]);
      ckmax(dp[u][i + 1],
            dp[ls][i] + 1ll * (m - 1) * a[pos] - 2ll * a[pos] * i);
    }
  }
  if (rs) {
    for (int i = 0; i <= sz[rs]; i++) {
      ckmax(dp[u][i], dp[rs][i]);
      ckmax(dp[u][i + 1],
            dp[rs][i] + 1ll * (m - 1) * a[pos] - 2ll * a[pos] * i);
    }
  }
}
int main() {
  memset(dp, -0x3f, sizeof(dp));
  n = read(), m = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  int rt;
  solve(rt, 1, n);
  print(dp[rt][m], '\n');
  return 0;
}

### Prompt

Your challenge is to write 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>
#pragma GCC optimize(2, 3, "Ofast")
using namespace std;
template <typename T1, typename T2>
void ckmin(T1 &a, T2 b) {
  if (a > b) a = b;
}
template <typename T1, typename T2>
void ckmax(T1 &a, T2 b) {
  if (a < b) a = b;
}
int read() {
  int x = 0, f = 0;
  char ch = getchar();
  while (!isdigit(ch)) f |= ch == '-', ch = getchar();
  while (isdigit(ch)) x = 10 * x + ch - '0', ch = getchar();
  return f ? -x : x;
}
template <typename T>
void print(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) print(x / 10);
  putchar(x % 10 + '0');
}
template <typename T>
void print(T x, char let) {
  print(x), putchar(let);
}
const int N = 4005;
long long dp[N][N];
int sz[N], a[N], n, m, tot;
void solve(int &u, int l, int r) {
  if (l > r) {
    u = 0;
    return;
  }
  u = ++tot, sz[u] = r - l + 1;
  dp[u][0] = 0;
  dp[u][1] = 1ll * (m - 1) * a[l];
  int pos = l, ls = 0, rs = 0;
  for (int i = l; i <= r; i++)
    if (a[i] < a[pos]) pos = i;
  solve(ls, l, pos - 1), solve(rs, pos + 1, r);
  if (ls && rs) {
    for (int i = 1; i <= sz[ls]; i++) {
      for (int j = 1; j <= sz[rs]; j++) {
        ckmax(dp[u][i + j], dp[ls][i] + dp[rs][j] - 2ll * a[pos] * i * j);
        ckmax(dp[u][i + j + 1], dp[ls][i] + dp[rs][j] + 1ll * (m - 1) * a[pos] -
                                    2ll * a[pos] * (1ll * i * j + i + j));
      }
    }
  }
  if (ls) {
    for (int i = 0; i <= sz[ls]; i++) {
      ckmax(dp[u][i], dp[ls][i]);
      ckmax(dp[u][i + 1],
            dp[ls][i] + 1ll * (m - 1) * a[pos] - 2ll * a[pos] * i);
    }
  }
  if (rs) {
    for (int i = 0; i <= sz[rs]; i++) {
      ckmax(dp[u][i], dp[rs][i]);
      ckmax(dp[u][i + 1],
            dp[rs][i] + 1ll * (m - 1) * a[pos] - 2ll * a[pos] * i);
    }
  }
}
int main() {
  memset(dp, -0x3f, sizeof(dp));
  n = read(), m = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  int rt;
  solve(rt, 1, n);
  print(dp[rt][m], '\n');
  return 0;
}

```