#include <bits/stdc++.h>
using namespace std;
int n, m;
long long a[4005];
int ls[4005], rs[4005], siz[4005];
vector<int> S;
void build() {
  for (int i = 1; i <= n; i++) {
    int cur = 0;
    while (S.size() && a[S.back()] >= a[i]) {
      cur = S.back();
      S.pop_back();
    }
    if (S.size()) rs[S.back()] = i;
    ls[i] = cur;
    S.push_back(i);
  }
}
long long dp[4005][4005];
long long Max(long long a, long long b) {
  if (a > b) return a;
  return b;
}
void dfs1(int u) {
  for (int i = 1; i <= n; i++) dp[u][i] = -114514191981011451;
  siz[u] = 1;
  if (ls[u]) dfs1(ls[u]);
  siz[u] += siz[ls[u]];
  if (rs[u]) dfs1(rs[u]);
  siz[u] += siz[rs[u]];
  for (int i = 0; i <= siz[ls[u]]; i++) {
    for (int j = 0; j <= siz[rs[u]]; j++) {
      dp[u][i + j] =
          Max(dp[u][i + j], dp[ls[u]][i] + dp[rs[u]][j] - a[u] * 2 * i * j);
    }
  }
  for (int i = siz[u]; i >= 1; i--)
    dp[u][i] = Max(dp[u][i - 1] + a[u] * m - a[u] * (2 * i - 1), dp[u][i]);
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  build();
  dfs1(S[0]);
  printf("%lld\n", dp[S[0]][m]);
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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;
long long a[4005];
int ls[4005], rs[4005], siz[4005];
vector<int> S;
void build() {
  for (int i = 1; i <= n; i++) {
    int cur = 0;
    while (S.size() && a[S.back()] >= a[i]) {
      cur = S.back();
      S.pop_back();
    }
    if (S.size()) rs[S.back()] = i;
    ls[i] = cur;
    S.push_back(i);
  }
}
long long dp[4005][4005];
long long Max(long long a, long long b) {
  if (a > b) return a;
  return b;
}
void dfs1(int u) {
  for (int i = 1; i <= n; i++) dp[u][i] = -114514191981011451;
  siz[u] = 1;
  if (ls[u]) dfs1(ls[u]);
  siz[u] += siz[ls[u]];
  if (rs[u]) dfs1(rs[u]);
  siz[u] += siz[rs[u]];
  for (int i = 0; i <= siz[ls[u]]; i++) {
    for (int j = 0; j <= siz[rs[u]]; j++) {
      dp[u][i + j] =
          Max(dp[u][i + j], dp[ls[u]][i] + dp[rs[u]][j] - a[u] * 2 * i * j);
    }
  }
  for (int i = siz[u]; i >= 1; i--)
    dp[u][i] = Max(dp[u][i - 1] + a[u] * m - a[u] * (2 * i - 1), dp[u][i]);
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  build();
  dfs1(S[0]);
  printf("%lld\n", dp[S[0]][m]);
  return 0;
}

```