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>
  inline void read(_T &x) {
    x = 0;
    long long f = 1;
    char s = getchar();
    while (s < '0' || s > '9') {
      f = 1;
      if (s == '-') f = -1;
      s = getchar();
    }
    while ('0' <= s && s <= '9') {
      x = (x << 3) + (x << 1) + s - '0';
      s = getchar();
    }
    x *= f;
  }
  const long long np = 1e5 + 5;
  const long long INF = 1 << 30;
  long long val[np];
  long long dp[4005][4005];
  long long ls[np], rs[np];
  long long sta[np], top;
  long long siz[np];
  long long n, m, tmp[np];
  inline void dfs(long long x) {
    dp[x][0] = 0;
    dp[x][1] = m * val[x] - val[x];
    siz[x] = 1;
    if (!ls[x] && !rs[x]) return;
    if (ls[x]) {
      dfs(ls[x]);
      for (long long i = 0; i <= siz[x] + siz[ls[x]]; i++) tmp[i] = 0;
      for (long long i = 0; i <= siz[x]; i++) {
        for (long long j = 0; j <= siz[ls[x]]; j++) {
          tmp[i + j] =
              max(tmp[i + j], dp[x][i] + dp[ls[x]][j] - val[x] * 2 * i * j);
        }
      }
      siz[x] += siz[ls[x]];
      for (long long i = 0; i <= siz[x]; i++) dp[x][i] = tmp[i];
    }
    if (rs[x]) {
      dfs(rs[x]);
      for (long long i = 0; i <= siz[x] + siz[rs[x]]; i++) tmp[i] = 0;
      for (long long i = 0; i <= siz[x]; i++) {
        for (long long j = 0; j <= siz[rs[x]]; j++) {
          tmp[i + j] =
              max(tmp[i + j], dp[x][i] + dp[rs[x]][j] - val[x] * 2 * i * j);
        }
      }
      siz[x] += siz[rs[x]];
      for (long long i = 0; i <= siz[x]; i++) dp[x][i] = tmp[i];
    }
  }
  signed main() {
    dp[0][0] = 0;
    read(n), read(m);
    for (long long i = 1; i <= n; i++) read(val[i]);
    for (long long i = 1; i <= n; i++) {
      long long top_ = top;
      while (top && val[sta[top]] > val[i]) top--;
      if (top) rs[sta[top]] = i;
      if (top < top_) ls[i] = sta[top + 1];
      sta[++top] = i;
    }
    dfs(sta[1]);
    printf("%lld\n", dp[sta[1]][m]);
    return 0;
  }
  