#include <bits/stdc++.h>
int main(void) {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  std::cout.tie(nullptr);
  int N, M;
  std::cin >> N >> M;
  std::vector<int> A(N + 1, 0);
  for (int i = 1; i <= N; ++i) std::cin >> A[i];
  std::vector<int> Sta(N + 1, 0), Ls(N + 1, 0), Rs(N + 1, 0);
  int top = 0, mn = A[1];
  for (int i = 1; i <= N; ++i) {
    while (top && A[Sta[top]] > A[i]) Ls[i] = Sta[top--];
    if (top) Rs[Sta[top]] = i;
    Sta[++top] = i, mn = std::min(mn, A[i]);
  }
  int rt = 0;
  for (int i = 1; i <= N; ++i)
    if (mn == A[i]) rt = i;
  std::vector<std::vector<long long>> dp(N + 1,
                                         std::vector<long long>(M + 1, -1E17));
  std::vector<int> Sz(N + 1, 0);
  auto Merge = [&](int u, int v) {
    if (!v) return;
    std::vector<long long> f(M + 1, -1E18);
    for (int a = 0; a <= Sz[u]; ++a)
      for (int b = 0; b <= Sz[v] && a + b <= M; ++b)
        f[a + b] = std::max(f[a + b], dp[u][a] + dp[v][b] - 2ll * a * b * A[u]);
    dp[u] = f, Sz[u] += Sz[v];
  };
  std::function<void(int)> solve = [&](int u) {
    if (!u) return;
    dp[u][0] = 0, dp[u][1] = 1ll * A[u] * (M - 1), Sz[u] = 1;
    solve(Ls[u]), solve(Rs[u]);
    Merge(u, Ls[u]), Merge(u, Rs[u]);
  };
  solve(rt);
  long long ans = -2E18;
  for (int i = 1; i <= N; ++i) ans = std::max(ans, dp[i][M]);
  std::cout << ans << '\n';
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 main(void) {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  std::cout.tie(nullptr);
  int N, M;
  std::cin >> N >> M;
  std::vector<int> A(N + 1, 0);
  for (int i = 1; i <= N; ++i) std::cin >> A[i];
  std::vector<int> Sta(N + 1, 0), Ls(N + 1, 0), Rs(N + 1, 0);
  int top = 0, mn = A[1];
  for (int i = 1; i <= N; ++i) {
    while (top && A[Sta[top]] > A[i]) Ls[i] = Sta[top--];
    if (top) Rs[Sta[top]] = i;
    Sta[++top] = i, mn = std::min(mn, A[i]);
  }
  int rt = 0;
  for (int i = 1; i <= N; ++i)
    if (mn == A[i]) rt = i;
  std::vector<std::vector<long long>> dp(N + 1,
                                         std::vector<long long>(M + 1, -1E17));
  std::vector<int> Sz(N + 1, 0);
  auto Merge = [&](int u, int v) {
    if (!v) return;
    std::vector<long long> f(M + 1, -1E18);
    for (int a = 0; a <= Sz[u]; ++a)
      for (int b = 0; b <= Sz[v] && a + b <= M; ++b)
        f[a + b] = std::max(f[a + b], dp[u][a] + dp[v][b] - 2ll * a * b * A[u]);
    dp[u] = f, Sz[u] += Sz[v];
  };
  std::function<void(int)> solve = [&](int u) {
    if (!u) return;
    dp[u][0] = 0, dp[u][1] = 1ll * A[u] * (M - 1), Sz[u] = 1;
    solve(Ls[u]), solve(Rs[u]);
    Merge(u, Ls[u]), Merge(u, Rs[u]);
  };
  solve(rt);
  long long ans = -2E18;
  for (int i = 1; i <= N; ++i) ans = std::max(ans, dp[i][M]);
  std::cout << ans << '\n';
  return 0;
}

```