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;
  const long long MAXN = 10000000000000;
  long long n, a[4003], m, dp[4003][4003], lson[4003], rson[4003], siz[4003],
      lval[4003], rval[4003];
  void solve(long long l, long long r, long long pos) {
    if (l == r) return;
    long long mn, now;
    mn = now = MAXN;
    for (long long i = l; i < pos; ++i)
      if (a[i] < mn) mn = a[i], now = i;
    if (mn < MAXN) {
      lson[pos] = now;
      lval[pos] = mn - a[pos];
      solve(l, pos - 1, now);
    }
    mn = now = MAXN;
    for (long long i = pos + 1; i <= r; ++i)
      if (a[i] < mn) mn = a[i], now = i;
    if (mn < MAXN) {
      rson[pos] = now;
      rval[pos] = mn - a[pos];
      solve(pos + 1, r, now);
    }
  }
  void dfs(long long pos) {
    siz[pos] = 1;
    if (lson[pos]) {
      dfs(lson[pos]);
      for (long long i = siz[pos]; i >= 0; --i)
        for (long long j = siz[lson[pos]]; j >= 0; --j)
          dp[pos][i + j] = max(dp[pos][i + j], dp[lson[pos]][j] + dp[pos][i] +
                                                   j * (m - j) * lval[pos]);
      siz[pos] += siz[lson[pos]];
    }
    if (rson[pos]) {
      dfs(rson[pos]);
      for (long long i = siz[pos]; i >= 0; --i)
        for (long long j = siz[rson[pos]]; j >= 0; --j)
          dp[pos][i + j] = max(dp[pos][i + j], dp[rson[pos]][j] + dp[pos][i] +
                                                   j * (m - j) * rval[pos]);
      siz[pos] += siz[rson[pos]];
    }
  }
  signed main() {
    cin >> n >> m;
    for (long long i = 1; i <= n; ++i) cin >> a[i];
    long long to, mn = MAXN;
    for (long long i = 1; i <= n; ++i)
      if (a[i] < mn) mn = a[i], to = i;
    solve(1, n, to);
    dfs(to);
    printf("%lld\n", dp[to][m]);
    return 0;
  }
  