#include <bits/stdc++.h>
using namespace std;
const long long MAXN = 4010;
const long long INF = 2e9;
const double eps = 1e-6;
const double PI = acos(-1);
long long n, m, rt, a[MAXN];
long long p[MAXN], ls[MAXN], rs[MAXN], siz[MAXN];
long long stk[MAXN], top, dp[MAXN][MAXN];
template <typename T>
inline bool read(T &a) {
  a = 0;
  char c = getchar();
  long long f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    a = a * 10 + (c ^ 48);
    c = getchar();
  }
  return a *= f, true;
}
template <typename A, typename... B>
inline bool read(A &x, B &...y) {
  return read(x) && read(y...);
}
void DFS(long long x) {
  siz[x] = 1, dp[x][0] = 0, dp[x][1] = m * a[x] - a[x];
  if (ls[x]) {
    DFS(ls[x]);
    for (long long i = siz[x]; i >= 0; --i) {
      for (long long j = 1; j <= siz[ls[x]]; ++j)
        dp[x][i + j] =
            max(dp[x][i + j], dp[x][i] + dp[ls[x]][j] - 2 * i * j * a[x]);
    }
    siz[x] += siz[ls[x]];
  }
  if (rs[x]) {
    DFS(rs[x]);
    for (long long i = siz[x]; i >= 0; --i) {
      for (long long j = 1; j <= siz[rs[x]]; ++j)
        dp[x][i + j] =
            max(dp[x][i + j], dp[x][i] + dp[rs[x]][j] - 2 * i * j * a[x]);
    }
    siz[x] += siz[rs[x]];
  }
}
signed main() {
  read(n), read(m);
  for (long long i = 1; i <= n; ++i) read(a[i]);
  for (long long i = 1; i <= n; ++i) {
    while (top && a[stk[top]] > a[i]) ls[i] = stk[top--];
    if (top) rs[stk[top]] = i;
    stk[++top] = i;
  }
  rt = stk[1];
  DFS(rt);
  printf("%lld\n", dp[rt][m]);
  return 0;
}

### Prompt

Create a solution in Cpp for the following problem:

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 MAXN = 4010;
const long long INF = 2e9;
const double eps = 1e-6;
const double PI = acos(-1);
long long n, m, rt, a[MAXN];
long long p[MAXN], ls[MAXN], rs[MAXN], siz[MAXN];
long long stk[MAXN], top, dp[MAXN][MAXN];
template <typename T>
inline bool read(T &a) {
  a = 0;
  char c = getchar();
  long long f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    a = a * 10 + (c ^ 48);
    c = getchar();
  }
  return a *= f, true;
}
template <typename A, typename... B>
inline bool read(A &x, B &...y) {
  return read(x) && read(y...);
}
void DFS(long long x) {
  siz[x] = 1, dp[x][0] = 0, dp[x][1] = m * a[x] - a[x];
  if (ls[x]) {
    DFS(ls[x]);
    for (long long i = siz[x]; i >= 0; --i) {
      for (long long j = 1; j <= siz[ls[x]]; ++j)
        dp[x][i + j] =
            max(dp[x][i + j], dp[x][i] + dp[ls[x]][j] - 2 * i * j * a[x]);
    }
    siz[x] += siz[ls[x]];
  }
  if (rs[x]) {
    DFS(rs[x]);
    for (long long i = siz[x]; i >= 0; --i) {
      for (long long j = 1; j <= siz[rs[x]]; ++j)
        dp[x][i + j] =
            max(dp[x][i + j], dp[x][i] + dp[rs[x]][j] - 2 * i * j * a[x]);
    }
    siz[x] += siz[rs[x]];
  }
}
signed main() {
  read(n), read(m);
  for (long long i = 1; i <= n; ++i) read(a[i]);
  for (long long i = 1; i <= n; ++i) {
    while (top && a[stk[top]] > a[i]) ls[i] = stk[top--];
    if (top) rs[stk[top]] = i;
    stk[++top] = i;
  }
  rt = stk[1];
  DFS(rt);
  printf("%lld\n", dp[rt][m]);
  return 0;
}

```