#include <bits/stdc++.h>
int n, m, ls[4005], rs[4005];
long long a[4005], siz[4005], dp[4005][4005];
std::vector<int> st;
void cmax(long long &A, long long B) { A = std::max(A, B); }
void dfs(int now) {
  siz[now] = 1;
  if (ls[now]) dfs(ls[now]), siz[now] += siz[ls[now]];
  if (rs[now]) dfs(rs[now]), siz[now] += siz[rs[now]];
  if (ls[now] && rs[now]) {
    for (int i = 0; i <= siz[ls[now]]; ++i)
      for (int j = 0; j <= siz[rs[now]]; ++j) {
        cmax(dp[now][i + j],
             dp[ls[now]][i] + dp[rs[now]][j] - 2 * a[now] * i * j);
        cmax(dp[now][i + j + 1], dp[ls[now]][i] + dp[rs[now]][j] +
                                     a[now] * (m - 1 - 2 * (i + j)) -
                                     2 * a[now] * i * j);
      }
  } else if (ls[now]) {
    for (int i = 0; i <= siz[ls[now]]; ++i) {
      cmax(dp[now][i], dp[ls[now]][i]);
      cmax(dp[now][i + 1], dp[ls[now]][i] + a[now] * (m - 1 - 2 * i));
    }
  } else if (rs[now]) {
    for (int i = 0; i <= siz[rs[now]]; ++i) {
      cmax(dp[now][i], dp[rs[now]][i]);
      cmax(dp[now][i + 1], dp[rs[now]][i] + a[now] * (m - 1 - 2 * i));
    }
  } else {
    dp[now][0] = 0;
    dp[now][1] = a[now] * (m - 1);
  }
}
signed main() {
  std::ios::sync_with_stdio(false);
  std::cin >> n >> m;
  st.push_back(0);
  for (int i = 1; i <= n; ++i) {
    std::cin >> a[i];
    while (!st.empty() && a[st.back()] > a[i]) ls[i] = st.back(), st.pop_back();
    rs[st.back()] = i;
    st.push_back(i);
  }
  memset(dp, 0xf0, sizeof(dp));
  dfs(rs[0]);
  std::cout << dp[rs[0]][m] << std::endl;
  return 0;
}

### Prompt

Please formulate 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>
int n, m, ls[4005], rs[4005];
long long a[4005], siz[4005], dp[4005][4005];
std::vector<int> st;
void cmax(long long &A, long long B) { A = std::max(A, B); }
void dfs(int now) {
  siz[now] = 1;
  if (ls[now]) dfs(ls[now]), siz[now] += siz[ls[now]];
  if (rs[now]) dfs(rs[now]), siz[now] += siz[rs[now]];
  if (ls[now] && rs[now]) {
    for (int i = 0; i <= siz[ls[now]]; ++i)
      for (int j = 0; j <= siz[rs[now]]; ++j) {
        cmax(dp[now][i + j],
             dp[ls[now]][i] + dp[rs[now]][j] - 2 * a[now] * i * j);
        cmax(dp[now][i + j + 1], dp[ls[now]][i] + dp[rs[now]][j] +
                                     a[now] * (m - 1 - 2 * (i + j)) -
                                     2 * a[now] * i * j);
      }
  } else if (ls[now]) {
    for (int i = 0; i <= siz[ls[now]]; ++i) {
      cmax(dp[now][i], dp[ls[now]][i]);
      cmax(dp[now][i + 1], dp[ls[now]][i] + a[now] * (m - 1 - 2 * i));
    }
  } else if (rs[now]) {
    for (int i = 0; i <= siz[rs[now]]; ++i) {
      cmax(dp[now][i], dp[rs[now]][i]);
      cmax(dp[now][i + 1], dp[rs[now]][i] + a[now] * (m - 1 - 2 * i));
    }
  } else {
    dp[now][0] = 0;
    dp[now][1] = a[now] * (m - 1);
  }
}
signed main() {
  std::ios::sync_with_stdio(false);
  std::cin >> n >> m;
  st.push_back(0);
  for (int i = 1; i <= n; ++i) {
    std::cin >> a[i];
    while (!st.empty() && a[st.back()] > a[i]) ls[i] = st.back(), st.pop_back();
    rs[st.back()] = i;
    st.push_back(i);
  }
  memset(dp, 0xf0, sizeof(dp));
  dfs(rs[0]);
  std::cout << dp[rs[0]][m] << std::endl;
  return 0;
}

```