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>
  long long read(long long x = 0, long long f = 0, char ch = getchar()) {
    while (ch < 48 or ch > 57) f = ch == 45, ch = getchar();
    while (48 <= ch and ch <= 57) x = x * 10 + ch - 48, ch = getchar();
    return f ? -x : x;
  }
  const long long N = 1e6 + 5;
  long long n, m, a[N];
  std::vector<long long> solve(long long L, long long R) {
    std::vector<long long> res(R - L + 2, -1LL << 60);
    res[0] = 0;
    if (L > R) return res;
    long long pos = std::min_element(a + L, a + R + 1) - a;
    std::vector<long long> x = solve(L, pos - 1), y = solve(pos + 1, R);
    for (long long i = 0; i < x.size(); i++)
      for (long long j = 0; j < y.size(); j++) {
        res[i + j] = std::max(res[i + j], x[i] + y[j] - 2 * i * j * a[pos]);
        res[i + j + 1] = std::max(
            res[i + j + 1],
            x[i] + y[j] - (2 * (i + 1) * (j + 1) - 1) * a[pos] + m * a[pos]);
      }
    return res;
  }
  signed main() {
    n = read(), m = read();
    for (long long i = 1; i <= n; i++) a[i] = read();
    printf("%lld", solve(1, n)[m]);
    return 0;
  }
  