#include <bits/stdc++.h>
using namespace std;
const long long S = 1 << 22;
char p0[S], *p1, *p2;
inline long long read() {
  long long x = 0, f = 0;
  char c =
      (p2 == p1 && (p2 = (p1 = p0) + fread(p0, 1, S, stdin)) == p1 ? EOF
                                                                   : *p1++);
  while (!isdigit(c)) {
    if (c == '-') f = 1;
    c = (p2 == p1 && (p2 = (p1 = p0) + fread(p0, 1, S, stdin)) == p1 ? EOF
                                                                     : *p1++);
  }
  while (isdigit(c))
    x = (x << 1) + (x << 3) + (c & 15),
    c = (p2 == p1 && (p2 = (p1 = p0) + fread(p0, 1, S, stdin)) == p1 ? EOF
                                                                     : *p1++);
  return f ? -x : x;
}
const long long MAXN = 4000;
long long a[MAXN + 10], st[MAXN + 10], top = 0;
long long lc[MAXN + 10], rc[MAXN + 10];
long long n, m;
long long dp[MAXN + 10][MAXN + 10], siz[MAXN + 10];
void dfs(long long x) {
  if (lc[x]) dfs(lc[x]);
  if (rc[x]) dfs(rc[x]);
  siz[x] = siz[lc[x]] + siz[rc[x]] + 1;
  for (long long i = 0; i <= siz[lc[x]]; ++i) {
    for (long long j = 0; j <= siz[rc[x]]; ++j) {
      dp[x][i + j] =
          max(dp[x][i + j], dp[lc[x]][i] + dp[rc[x]][j] - 2 * a[x] * (i * j));
      dp[x][i + j + 1] = max(
          dp[x][i + j + 1], dp[lc[x]][i] + dp[rc[x]][j] - 2 * a[x] * (i) * (j) +
                                m * a[x] - 2 * a[x] * (i + j) - a[x]);
    }
  }
}
signed main() {
  n = read(), m = read();
  for (long long i = 1; i <= n; ++i) a[i] = read();
  for (long long i = 1; i <= n; ++i) {
    long long k = top;
    while (k && a[st[k]] > a[i]) k--;
    if (k) rc[st[k]] = i;
    if (k < top) lc[i] = st[k + 1];
    st[++k] = i;
    top = k;
  }
  dfs(st[1]);
  long long ans = dp[st[1]][m];
  ;
  cout << ans << endl;
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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].

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long S = 1 << 22;
char p0[S], *p1, *p2;
inline long long read() {
  long long x = 0, f = 0;
  char c =
      (p2 == p1 && (p2 = (p1 = p0) + fread(p0, 1, S, stdin)) == p1 ? EOF
                                                                   : *p1++);
  while (!isdigit(c)) {
    if (c == '-') f = 1;
    c = (p2 == p1 && (p2 = (p1 = p0) + fread(p0, 1, S, stdin)) == p1 ? EOF
                                                                     : *p1++);
  }
  while (isdigit(c))
    x = (x << 1) + (x << 3) + (c & 15),
    c = (p2 == p1 && (p2 = (p1 = p0) + fread(p0, 1, S, stdin)) == p1 ? EOF
                                                                     : *p1++);
  return f ? -x : x;
}
const long long MAXN = 4000;
long long a[MAXN + 10], st[MAXN + 10], top = 0;
long long lc[MAXN + 10], rc[MAXN + 10];
long long n, m;
long long dp[MAXN + 10][MAXN + 10], siz[MAXN + 10];
void dfs(long long x) {
  if (lc[x]) dfs(lc[x]);
  if (rc[x]) dfs(rc[x]);
  siz[x] = siz[lc[x]] + siz[rc[x]] + 1;
  for (long long i = 0; i <= siz[lc[x]]; ++i) {
    for (long long j = 0; j <= siz[rc[x]]; ++j) {
      dp[x][i + j] =
          max(dp[x][i + j], dp[lc[x]][i] + dp[rc[x]][j] - 2 * a[x] * (i * j));
      dp[x][i + j + 1] = max(
          dp[x][i + j + 1], dp[lc[x]][i] + dp[rc[x]][j] - 2 * a[x] * (i) * (j) +
                                m * a[x] - 2 * a[x] * (i + j) - a[x]);
    }
  }
}
signed main() {
  n = read(), m = read();
  for (long long i = 1; i <= n; ++i) a[i] = read();
  for (long long i = 1; i <= n; ++i) {
    long long k = top;
    while (k && a[st[k]] > a[i]) k--;
    if (k) rc[st[k]] = i;
    if (k < top) lc[i] = st[k + 1];
    st[++k] = i;
    top = k;
  }
  dfs(st[1]);
  long long ans = dp[st[1]][m];
  ;
  cout << ans << endl;
  return 0;
}

```