#include <bits/stdc++.h>
using namespace std;
using ll = long long;
constexpr int maxn = 4e3 + 7;
int siz[maxn], ls[maxn], rs[maxn], lw[maxn], rw[maxn], arr[maxn];
int sta[maxn], n, m;
ll dp[maxn][maxn];
void build() {
  int top = 0;
  for (int i = 1; i <= n; ++i) {
    while (top > 0 && arr[sta[top]] > arr[i]) ls[i] = sta[top--];
    if (sta[top]) {
      rs[sta[top]] = i;
      rw[sta[top]] = arr[i] - arr[sta[top]];
    }
    if (ls[i]) {
      lw[i] = arr[ls[i]] - arr[i];
    }
    sta[++top] = i;
  }
}
void dfs(int u) {
  siz[u] = 1;
  if (ls[u]) {
    dfs(ls[u]);
    for (int i = min(m, siz[u]); i >= 0; --i) {
      for (int j = min(m, siz[ls[u]]); j >= 0; --j) {
        if (i + j > m) continue;
        dp[u][i + j] = max(dp[u][i + j],
                           dp[u][i] + dp[ls[u]][j] + 1ll * (m - j) * j * lw[u]);
      }
    }
    siz[u] += siz[ls[u]];
  }
  if (rs[u]) {
    dfs(rs[u]);
    for (int i = min(m, siz[u]); i >= 0; --i) {
      for (int j = min(m, siz[rs[u]]); j >= 0; --j) {
        if (i + j > m) continue;
        dp[u][i + j] = max(dp[u][i + j],
                           dp[u][i] + dp[rs[u]][j] + 1ll * (m - j) * j * rw[u]);
      }
    }
    siz[u] += siz[rs[u]];
  }
}
void solve() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d", arr + i);
  build();
  dfs(sta[1]);
  printf("%lld\n", dp[sta[1]][m]);
}
int main() {
  solve();
  return 0;
}

### Prompt

Generate 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>
using namespace std;
using ll = long long;
constexpr int maxn = 4e3 + 7;
int siz[maxn], ls[maxn], rs[maxn], lw[maxn], rw[maxn], arr[maxn];
int sta[maxn], n, m;
ll dp[maxn][maxn];
void build() {
  int top = 0;
  for (int i = 1; i <= n; ++i) {
    while (top > 0 && arr[sta[top]] > arr[i]) ls[i] = sta[top--];
    if (sta[top]) {
      rs[sta[top]] = i;
      rw[sta[top]] = arr[i] - arr[sta[top]];
    }
    if (ls[i]) {
      lw[i] = arr[ls[i]] - arr[i];
    }
    sta[++top] = i;
  }
}
void dfs(int u) {
  siz[u] = 1;
  if (ls[u]) {
    dfs(ls[u]);
    for (int i = min(m, siz[u]); i >= 0; --i) {
      for (int j = min(m, siz[ls[u]]); j >= 0; --j) {
        if (i + j > m) continue;
        dp[u][i + j] = max(dp[u][i + j],
                           dp[u][i] + dp[ls[u]][j] + 1ll * (m - j) * j * lw[u]);
      }
    }
    siz[u] += siz[ls[u]];
  }
  if (rs[u]) {
    dfs(rs[u]);
    for (int i = min(m, siz[u]); i >= 0; --i) {
      for (int j = min(m, siz[rs[u]]); j >= 0; --j) {
        if (i + j > m) continue;
        dp[u][i + j] = max(dp[u][i + j],
                           dp[u][i] + dp[rs[u]][j] + 1ll * (m - j) * j * rw[u]);
      }
    }
    siz[u] += siz[rs[u]];
  }
}
void solve() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d", arr + i);
  build();
  dfs(sta[1]);
  printf("%lld\n", dp[sta[1]][m]);
}
int main() {
  solve();
  return 0;
}

```