description: 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].
solution: #include <bits/stdc++.h>
  using namespace std;
  template <typename T>
  bool chkmax(T &x, T y) {
    return x < y ? x = y, true : false;
  }
  template <typename T>
  bool chkmin(T &x, T y) {
    return x > y ? x = y, true : false;
  }
  int readint() {
    int x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
      x = x * 10 + ch - '0';
      ch = getchar();
    }
    return x * f;
  }
  int n, m, rt;
  long long a[4005], rmq[4005][20], ch[4005][2], lg2[4005], siz[4005];
  long long d[4005][4005];
  int getmin(int l, int r) {
    int k = lg2[r - l + 1];
    return a[rmq[l][k]] < a[rmq[r - (1 << k) + 1][k]] ? rmq[l][k]
                                                      : rmq[r - (1 << k) + 1][k];
  }
  int build(int l, int r) {
    int mid = getmin(l, r);
    if (l != mid) ch[mid][0] = build(l, mid - 1);
    if (r != mid) ch[mid][1] = build(mid + 1, r);
    return mid;
  }
  void dfs(int u) {
    if (!u) return;
    siz[u] = 1;
    dfs(ch[u][0]), dfs(ch[u][1]);
    siz[u] += siz[ch[u][0]] + siz[ch[u][1]];
    for (int i = 0; i <= siz[ch[u][0]]; i++) {
      for (int j = 0; j <= siz[ch[u][1]]; j++) {
        chkmax(d[u][i + j], d[ch[u][0]][i] + d[ch[u][1]][j] - 2ll * i * j * a[u]);
        chkmax(d[u][i + j + 1], d[ch[u][0]][i] + d[ch[u][1]][j] -
                                    2ll * i * j * a[u] + 1ll * (m - 1) * a[u] -
                                    2ll * (i + j) * a[u]);
      }
    }
  }
  int main() {
    n = readint();
    m = readint();
    for (int i = 1; i <= n; i++) a[i] = readint();
    for (int i = 2; i <= n; i++) lg2[i] = lg2[i >> 1] + 1;
    for (int i = 1; i <= n; i++) rmq[i][0] = i;
    for (int i = 1; (1 << i) <= n; i++)
      for (int j = 1; j + (1 << i) - 1 <= n; j++)
        rmq[j][i] = a[rmq[j][i - 1]] < a[rmq[j + (1 << (i - 1))][i - 1]]
                        ? rmq[j][i - 1]
                        : rmq[j + (1 << (i - 1))][i - 1];
    rt = build(1, n);
    for (int i = 0; i <= n; i++)
      for (int j = 0; j <= n; j++) d[i][j] = -(1ll << 60);
    d[0][0] = 0;
    dfs(rt);
    printf("%lld\n", d[rt][m]);
    return 0;
  }
  