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 ckmax(T &x, T y) {
    x = x < y ? y : x;
    return;
  }
  template <typename T>
  inline void ckmin(T &x, T y) {
    x = x < y ? x : y;
    return;
  }
  namespace FastIO {
  inline long long read() {
    long long res = 0, f = 1;
    char k;
    while (!isdigit(k = getchar()))
      if (k == '-') f = -1;
    while (isdigit(k)) res = res * 10 + k - '0', k = getchar();
    return res * f;
  }
  char OPUT[100];
  inline void print(long long x) {
    if (!x) return putchar('0'), putchar('\n'), void();
    if (x < 0) putchar('-'), x = -x;
    long long cnt = 0;
    while (x) OPUT[++cnt] = x % 10, x /= 10;
    for (register long long i = cnt; i >= 1; --i) putchar(OPUT[i] + '0');
    putchar('\n');
    return;
  }
  }  // namespace FastIO
  using namespace FastIO;
  const long long N = 4010;
  long long ls[N], rs[N], tmp[N], dp[N][N], a[N], siz[N], n, m;
  vector<long long> g[N];
  inline void solve(long long x) {
    siz[x] = 1;
    for (long long t : {ls[x], rs[x]})
      if (t) {
        solve(t);
        for (long long i = 1; i <= min(m, siz[x]); ++i) tmp[i] = dp[x][i];
        long long upi = min(siz[x], m), upj = min(siz[t], m);
        for (long long i = 0; i <= upi; ++i)
          for (long long j = 1; i + j <= m && j <= upj; ++j)
            ckmax(tmp[i + j], dp[x][i] + dp[t][j] + (a[t] - a[x]) * j * (m - j));
        siz[x] += siz[t];
        for (long long i = 1; i <= min(m, siz[x]); ++i)
          dp[x][i] = tmp[i], tmp[i] = 0;
      }
    return;
  }
  signed main() {
    n = read();
    m = read();
    for (register long long i = 1; i <= n; ++i) a[i] = read();
    stack<long long> st;
    for (long long i = 1; i <= n; ++i) {
      while (st.size() && a[st.top()] > a[i]) ls[i] = st.top(), st.pop();
      if (st.size()) rs[st.top()] = i;
      st.push(i);
    }
    while (st.size() > 1) st.pop();
    solve(st.top());
    print(dp[st.top()][m]);
    return 0;
  }
  