#include <bits/stdc++.h>
using namespace std;
const int N = 4005;
int pos[N][N], ls[N], rs[N], siz[N];
int build(int l, int r) {
  if (l == r) return l;
  int x = pos[l][r];
  if (x - 1 >= l) ls[x] = build(l, x - 1);
  if (x + 1 <= r) rs[x] = build(x + 1, r);
  return x;
}
int n, m;
long long a[N], dp[N][N];
void dfs(int x) {
  siz[x] = 1;
  if (ls[x]) dfs(ls[x]), siz[x] += siz[ls[x]];
  if (rs[x]) dfs(rs[x]), siz[x] += siz[rs[x]];
  if ((!ls[x]) && (!rs[x])) return;
  if (!ls[x]) {
    for (int i = 0; i <= siz[rs[x]]; i++) {
      dp[x][i] = max(dp[x][i], dp[rs[x]][i] + (a[rs[x]] - a[x]) * i * (m - i));
      dp[x][i + 1] =
          max(dp[x][i + 1], dp[rs[x]][i] + (a[rs[x]] - a[x]) * i * (m - i));
    }
    return;
  }
  if (!rs[x]) {
    for (int i = 0; i <= siz[ls[x]]; i++) {
      dp[x][i] = max(dp[x][i], dp[ls[x]][i] + (a[ls[x]] - a[x]) * i * (m - i));
      dp[x][i + 1] =
          max(dp[x][i + 1], dp[ls[x]][i] + (a[ls[x]] - a[x]) * i * (m - i));
    }
    return;
  }
  for (int i = 0; i <= siz[ls[x]]; i++)
    for (int j = 0; j <= siz[rs[x]]; j++) {
      long long add =
          (a[ls[x]] - a[x]) * i * (m - i) + (a[rs[x]] - a[x]) * j * (m - j);
      dp[x][i + j] = max(dp[x][i + j], dp[ls[x]][i] + dp[rs[x]][j] + add);
      dp[x][i + j + 1] =
          max(dp[x][i + j + 1], dp[ls[x]][i] + dp[rs[x]][j] + add);
    }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) {
    pos[i][i] = i;
    for (int j = i + 1; j <= n; j++)
      pos[i][j] = a[pos[i][j - 1]] < a[j] ? pos[i][j - 1] : j;
  }
  build(1, n);
  dfs(pos[1][n]);
  printf("%lld", dp[pos[1][n]][m]);
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
int pos[N][N], ls[N], rs[N], siz[N];
int build(int l, int r) {
  if (l == r) return l;
  int x = pos[l][r];
  if (x - 1 >= l) ls[x] = build(l, x - 1);
  if (x + 1 <= r) rs[x] = build(x + 1, r);
  return x;
}
int n, m;
long long a[N], dp[N][N];
void dfs(int x) {
  siz[x] = 1;
  if (ls[x]) dfs(ls[x]), siz[x] += siz[ls[x]];
  if (rs[x]) dfs(rs[x]), siz[x] += siz[rs[x]];
  if ((!ls[x]) && (!rs[x])) return;
  if (!ls[x]) {
    for (int i = 0; i <= siz[rs[x]]; i++) {
      dp[x][i] = max(dp[x][i], dp[rs[x]][i] + (a[rs[x]] - a[x]) * i * (m - i));
      dp[x][i + 1] =
          max(dp[x][i + 1], dp[rs[x]][i] + (a[rs[x]] - a[x]) * i * (m - i));
    }
    return;
  }
  if (!rs[x]) {
    for (int i = 0; i <= siz[ls[x]]; i++) {
      dp[x][i] = max(dp[x][i], dp[ls[x]][i] + (a[ls[x]] - a[x]) * i * (m - i));
      dp[x][i + 1] =
          max(dp[x][i + 1], dp[ls[x]][i] + (a[ls[x]] - a[x]) * i * (m - i));
    }
    return;
  }
  for (int i = 0; i <= siz[ls[x]]; i++)
    for (int j = 0; j <= siz[rs[x]]; j++) {
      long long add =
          (a[ls[x]] - a[x]) * i * (m - i) + (a[rs[x]] - a[x]) * j * (m - j);
      dp[x][i + j] = max(dp[x][i + j], dp[ls[x]][i] + dp[rs[x]][j] + add);
      dp[x][i + j + 1] =
          max(dp[x][i + j + 1], dp[ls[x]][i] + dp[rs[x]][j] + add);
    }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) {
    pos[i][i] = i;
    for (int j = i + 1; j <= n; j++)
      pos[i][j] = a[pos[i][j - 1]] < a[j] ? pos[i][j - 1] : j;
  }
  build(1, n);
  dfs(pos[1][n]);
  printf("%lld", dp[pos[1][n]][m]);
  return 0;
}

```