#include <bits/stdc++.h>
using namespace std;
int n, m;
int a[4010], size[4010];
int stk[4010], top;
int ls[4010], rs[4010], w[4010][2];
long long dp[4010][4010];
template <typename Tp>
void read(Tp &x) {
  x = 0;
  int f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  x *= f;
}
void dfs(int x) {
  size[x] = 1;
  if (ls[x]) {
    dfs(ls[x]);
    for (int i = size[x]; i >= 0; --i) {
      for (int j = size[ls[x]]; j >= 0; --j) {
        dp[x][i + j] = max(dp[x][i + j], dp[x][i] + dp[ls[x]][j] +
                                             1ll * j * (m - j) * w[x][0]);
      }
    }
    size[x] += size[ls[x]];
  }
  if (rs[x]) {
    dfs(rs[x]);
    for (int i = size[x]; i >= 0; --i) {
      for (int j = size[rs[x]]; j >= 0; --j) {
        dp[x][i + j] = max(dp[x][i + j], dp[x][i] + dp[rs[x]][j] +
                                             1ll * j * (m - j) * w[x][1]);
      }
    }
    size[x] += size[rs[x]];
  }
}
int main() {
  read(n), read(m);
  for (int i = 1; i <= n; ++i) {
    read(a[i]);
    bool flag = false;
    while (top > 0 && a[stk[top]] > a[i]) top--, flag = true;
    if (top) rs[stk[top]] = i, w[stk[top]][1] = a[i] - a[stk[top]];
    if (flag) ls[i] = stk[top + 1], w[i][0] = a[stk[top + 1]] - a[i];
    stk[++top] = i;
  }
  dfs(stk[1]);
  printf("%lld\n", dp[stk[1]][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;
int n, m;
int a[4010], size[4010];
int stk[4010], top;
int ls[4010], rs[4010], w[4010][2];
long long dp[4010][4010];
template <typename Tp>
void read(Tp &x) {
  x = 0;
  int f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  x *= f;
}
void dfs(int x) {
  size[x] = 1;
  if (ls[x]) {
    dfs(ls[x]);
    for (int i = size[x]; i >= 0; --i) {
      for (int j = size[ls[x]]; j >= 0; --j) {
        dp[x][i + j] = max(dp[x][i + j], dp[x][i] + dp[ls[x]][j] +
                                             1ll * j * (m - j) * w[x][0]);
      }
    }
    size[x] += size[ls[x]];
  }
  if (rs[x]) {
    dfs(rs[x]);
    for (int i = size[x]; i >= 0; --i) {
      for (int j = size[rs[x]]; j >= 0; --j) {
        dp[x][i + j] = max(dp[x][i + j], dp[x][i] + dp[rs[x]][j] +
                                             1ll * j * (m - j) * w[x][1]);
      }
    }
    size[x] += size[rs[x]];
  }
}
int main() {
  read(n), read(m);
  for (int i = 1; i <= n; ++i) {
    read(a[i]);
    bool flag = false;
    while (top > 0 && a[stk[top]] > a[i]) top--, flag = true;
    if (top) rs[stk[top]] = i, w[stk[top]][1] = a[i] - a[stk[top]];
    if (flag) ls[i] = stk[top + 1], w[i][0] = a[stk[top + 1]] - a[i];
    stk[++top] = i;
  }
  dfs(stk[1]);
  printf("%lld\n", dp[stk[1]][m]);
  return 0;
}

```