#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void read(T &a) {
  T w = 1;
  a = 0;
  char ch = getchar();
  for (; ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') w = -1;
  for (; ch >= '0' && ch <= '9'; ch = getchar()) a = (a * 10) + (ch - '0');
  a *= w;
}
template <typename T>
inline void ckmax(T &a, T b) {
  a = a > b ? a : b;
}
template <typename T>
inline void ckmin(T &a, T b) {
  a = a < b ? a : b;
}
const long long N = 4040 + 10;
long long a[N], ls[N], rs[N];
long long sta[N], top;
long long n, m;
long long siz[N];
long long dp[N][N];
void Dp(long long u) {
  siz[u] = 1;
  if (ls[u]) {
    Dp(ls[u]);
    siz[u] += siz[ls[u]];
  }
  if (rs[u]) {
    Dp(rs[u]);
    siz[u] += siz[rs[u]];
  }
  for (long long i = 0; i <= siz[ls[u]]; ++i) {
    for (long long j = 0; j <= siz[rs[u]]; ++j) {
      long long x = i + j;
      if (x > m) continue;
      ckmax(dp[u][x], dp[ls[u]][i] + dp[rs[u]][j] - 2ll * i * j * a[u]);
      ckmax(dp[u][x + 1], dp[ls[u]][i] + dp[rs[u]][j] - 2 * i * j * a[u] -
                              a[u] + m * a[u] - 2 * (i + j) * a[u]);
    }
  }
}
signed main() {
  read(n);
  read(m);
  long long mx = 0, rt = 1;
  for (long long i = 1; i <= n; ++i) {
    read(a[i]);
    if (i == 1) mx = a[i];
    while (top && a[sta[top]] > a[i]) top--;
    ls[i] = rs[sta[top]], rs[sta[top]] = i;
    sta[++top] = i;
    if (a[i] < mx) mx = a[i], rt = i;
  }
  Dp(rt);
  cout << dp[rt][m];
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
template <typename T>
inline void read(T &a) {
  T w = 1;
  a = 0;
  char ch = getchar();
  for (; ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') w = -1;
  for (; ch >= '0' && ch <= '9'; ch = getchar()) a = (a * 10) + (ch - '0');
  a *= w;
}
template <typename T>
inline void ckmax(T &a, T b) {
  a = a > b ? a : b;
}
template <typename T>
inline void ckmin(T &a, T b) {
  a = a < b ? a : b;
}
const long long N = 4040 + 10;
long long a[N], ls[N], rs[N];
long long sta[N], top;
long long n, m;
long long siz[N];
long long dp[N][N];
void Dp(long long u) {
  siz[u] = 1;
  if (ls[u]) {
    Dp(ls[u]);
    siz[u] += siz[ls[u]];
  }
  if (rs[u]) {
    Dp(rs[u]);
    siz[u] += siz[rs[u]];
  }
  for (long long i = 0; i <= siz[ls[u]]; ++i) {
    for (long long j = 0; j <= siz[rs[u]]; ++j) {
      long long x = i + j;
      if (x > m) continue;
      ckmax(dp[u][x], dp[ls[u]][i] + dp[rs[u]][j] - 2ll * i * j * a[u]);
      ckmax(dp[u][x + 1], dp[ls[u]][i] + dp[rs[u]][j] - 2 * i * j * a[u] -
                              a[u] + m * a[u] - 2 * (i + j) * a[u]);
    }
  }
}
signed main() {
  read(n);
  read(m);
  long long mx = 0, rt = 1;
  for (long long i = 1; i <= n; ++i) {
    read(a[i]);
    if (i == 1) mx = a[i];
    while (top && a[sta[top]] > a[i]) top--;
    ls[i] = rs[sta[top]], rs[sta[top]] = i;
    sta[++top] = i;
    if (a[i] < mx) mx = a[i], rt = i;
  }
  Dp(rt);
  cout << dp[rt][m];
  return 0;
}

```