#include <bits/stdc++.h>
using namespace std;
int n, m, rt, a[4005], st[4005], sz[4005], ch[4005][2];
long long dp[4005][4005];
void dfs(int u) {
  if (!u) return;
  dfs(ch[u][0]);
  dfs(ch[u][1]);
  sz[u] = sz[ch[u][0]] + sz[ch[u][1]] + 1;
  for (int i = 0; i <= sz[ch[u][0]]; ++i)
    for (int j = 0; j <= sz[ch[u][1]]; ++j)
      dp[u][i + j] = max(
          dp[u][i + j], dp[ch[u][0]][i] + dp[ch[u][1]][j] - 2ll * i * j * a[u]);
  for (int i = sz[u] - 1; i >= 0; --i)
    dp[u][i + 1] = max(dp[u][i + 1], dp[u][i] + 1ll * (m - i * 2 - 1) * a[u]);
}
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) {
    while (st[0] && a[i] < a[st[st[0]]]) ch[i][0] = st[st[0]--];
    ch[st[st[0]]][1] = i;
    st[++st[0]] = i;
  }
  dfs(st[1]);
  printf("%lld\n", dp[st[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, rt, a[4005], st[4005], sz[4005], ch[4005][2];
long long dp[4005][4005];
void dfs(int u) {
  if (!u) return;
  dfs(ch[u][0]);
  dfs(ch[u][1]);
  sz[u] = sz[ch[u][0]] + sz[ch[u][1]] + 1;
  for (int i = 0; i <= sz[ch[u][0]]; ++i)
    for (int j = 0; j <= sz[ch[u][1]]; ++j)
      dp[u][i + j] = max(
          dp[u][i + j], dp[ch[u][0]][i] + dp[ch[u][1]][j] - 2ll * i * j * a[u]);
  for (int i = sz[u] - 1; i >= 0; --i)
    dp[u][i + 1] = max(dp[u][i + 1], dp[u][i] + 1ll * (m - i * 2 - 1) * a[u]);
}
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) {
    while (st[0] && a[i] < a[st[st[0]]]) ch[i][0] = st[st[0]--];
    ch[st[st[0]]][1] = i;
    st[++st[0]] = i;
  }
  dfs(st[1]);
  printf("%lld\n", dp[st[1]][m]);
  return 0;
}

```