#include <bits/stdc++.h>
using namespace std;
const long long inf = 1000000000000000000;
const int N = 4005;
int n, m;
long long a[N], dp[N][N];
inline int read() {
  int F = 1, ANS = 0;
  char C = getchar();
  while (C < '0' || C > '9') {
    if (C == '-') F = -1;
    C = getchar();
  }
  while (C >= '0' && C <= '9') {
    ANS = ANS * 10 + C - '0';
    C = getchar();
  }
  return F * ANS;
}
int st[N], ls[N], rs[N], rt = 1;
long long lw[N], rw[N], size[N];
void build() {
  int top = 0, tot = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] < a[rt]) rt = i;
    tot = top;
    while (tot && a[st[tot]] > a[i]) tot--;
    if (tot) rs[st[tot]] = i, rw[st[tot]] = a[i] - a[st[tot]];
    if (tot < top) ls[i] = st[tot + 1], lw[i] = a[st[tot + 1]] - a[i];
    st[++tot] = i;
    top = tot;
  }
}
void dfs(int u) {
  size[u] = 1;
  if (ls[u]) {
    dfs(ls[u]);
    for (int i = size[u]; i >= 0; i--) {
      for (long long j = size[ls[u]]; j >= 0; j--) {
        dp[u][i + j] =
            max(dp[u][i + j], dp[u][i] + dp[ls[u]][j] + j * (m - j) * lw[u]);
      }
    }
    size[u] += size[ls[u]];
  }
  if (rs[u]) {
    dfs(rs[u]);
    for (int i = size[u]; i >= 0; i--) {
      for (long long j = size[rs[u]]; j >= 0; j--) {
        dp[u][i + j] =
            max(dp[u][i + j], dp[u][i] + dp[rs[u]][j] + j * (m - j) * rw[u]);
      }
    }
    size[u] += size[rs[u]];
  }
}
int main() {
  n = read(), m = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  build();
  dfs(rt);
  cout << dp[rt][m];
  return 0;
}

### Prompt

Please create a solution in CPP 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>
using namespace std;
const long long inf = 1000000000000000000;
const int N = 4005;
int n, m;
long long a[N], dp[N][N];
inline int read() {
  int F = 1, ANS = 0;
  char C = getchar();
  while (C < '0' || C > '9') {
    if (C == '-') F = -1;
    C = getchar();
  }
  while (C >= '0' && C <= '9') {
    ANS = ANS * 10 + C - '0';
    C = getchar();
  }
  return F * ANS;
}
int st[N], ls[N], rs[N], rt = 1;
long long lw[N], rw[N], size[N];
void build() {
  int top = 0, tot = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] < a[rt]) rt = i;
    tot = top;
    while (tot && a[st[tot]] > a[i]) tot--;
    if (tot) rs[st[tot]] = i, rw[st[tot]] = a[i] - a[st[tot]];
    if (tot < top) ls[i] = st[tot + 1], lw[i] = a[st[tot + 1]] - a[i];
    st[++tot] = i;
    top = tot;
  }
}
void dfs(int u) {
  size[u] = 1;
  if (ls[u]) {
    dfs(ls[u]);
    for (int i = size[u]; i >= 0; i--) {
      for (long long j = size[ls[u]]; j >= 0; j--) {
        dp[u][i + j] =
            max(dp[u][i + j], dp[u][i] + dp[ls[u]][j] + j * (m - j) * lw[u]);
      }
    }
    size[u] += size[ls[u]];
  }
  if (rs[u]) {
    dfs(rs[u]);
    for (int i = size[u]; i >= 0; i--) {
      for (long long j = size[rs[u]]; j >= 0; j--) {
        dp[u][i + j] =
            max(dp[u][i + j], dp[u][i] + dp[rs[u]][j] + j * (m - j) * rw[u]);
      }
    }
    size[u] += size[rs[u]];
  }
}
int main() {
  n = read(), m = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  build();
  dfs(rt);
  cout << dp[rt][m];
  return 0;
}

```