#include <bits/stdc++.h>
using namespace std;
template <typename T>
void chkMax(T &x, T y) {
  if (y > x) x = y;
}
template <typename T>
void chkMin(T &x, T y) {
  if (y < x) x = y;
}
template <typename T>
void inline read(T &x) {
  int f = 1;
  x = 0;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s <= '9' && s >= '0') x = x * 10 + (s ^ 48), s = getchar();
  x *= f;
}
const int N = 4005;
int n, m, L[N], R[N], s[N], top, sz[N], rt;
long long a[N], f[N][N];
void dfs(int u) {
  if (!u) return;
  sz[u] = 1;
  int ls = L[u], rs = R[u];
  dfs(ls);
  sz[u] += sz[ls];
  dfs(rs);
  sz[u] += sz[rs];
  for (int x = 0; x <= sz[ls]; x++) {
    for (int y = 0; y <= sz[rs]; y++) {
      chkMax(f[u][x + y], f[ls][x] + f[rs][y] - 2 * a[u] * x * y);
      chkMax(f[u][x + y + 1],
             f[ls][x] + f[rs][y] + (m - 1) * a[u] - 2 * a[u] * (x * y + x + y));
    }
  }
}
int main() {
  memset(f, 0xcf, sizeof f);
  f[0][0] = 0;
  read(n), read(m);
  for (int i = 1; i <= n; i++) read(a[i]);
  for (int i = 1; i <= n; i++) {
    while (top && a[s[top]] > a[i]) L[i] = s[top--];
    if (top) R[s[top]] = i;
    s[++top] = i;
  }
  rt = s[1];
  dfs(rt);
  printf("%lld\n", f[rt][m]);
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
template <typename T>
void chkMax(T &x, T y) {
  if (y > x) x = y;
}
template <typename T>
void chkMin(T &x, T y) {
  if (y < x) x = y;
}
template <typename T>
void inline read(T &x) {
  int f = 1;
  x = 0;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s <= '9' && s >= '0') x = x * 10 + (s ^ 48), s = getchar();
  x *= f;
}
const int N = 4005;
int n, m, L[N], R[N], s[N], top, sz[N], rt;
long long a[N], f[N][N];
void dfs(int u) {
  if (!u) return;
  sz[u] = 1;
  int ls = L[u], rs = R[u];
  dfs(ls);
  sz[u] += sz[ls];
  dfs(rs);
  sz[u] += sz[rs];
  for (int x = 0; x <= sz[ls]; x++) {
    for (int y = 0; y <= sz[rs]; y++) {
      chkMax(f[u][x + y], f[ls][x] + f[rs][y] - 2 * a[u] * x * y);
      chkMax(f[u][x + y + 1],
             f[ls][x] + f[rs][y] + (m - 1) * a[u] - 2 * a[u] * (x * y + x + y));
    }
  }
}
int main() {
  memset(f, 0xcf, sizeof f);
  f[0][0] = 0;
  read(n), read(m);
  for (int i = 1; i <= n; i++) read(a[i]);
  for (int i = 1; i <= n; i++) {
    while (top && a[s[top]] > a[i]) L[i] = s[top--];
    if (top) R[s[top]] = i;
    s[++top] = i;
  }
  rt = s[1];
  dfs(rt);
  printf("%lld\n", f[rt][m]);
  return 0;
}

```