#include <bits/stdc++.h>
using namespace std;
std::mt19937 rnd(
    (int)std::chrono::steady_clock::now().time_since_epoch().count());
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
const int MAXN = 4000;
int n, ntake;
int a[MAXN];
bool taken[MAXN];
long long cur[MAXN];
long long solve() {
  for (int i = (0); i < (n); ++i) taken[i] = false;
  for (int i = (0); i < (n); ++i) cur[i] = (long long)(ntake - 1) * a[i];
  long long ret = 0;
  for (int i = (0); i < (ntake); ++i) {
    int best = -1;
    for (int j = (0); j < (n); ++j)
      if (!taken[j] && (best == -1 || cur[j] > cur[best])) best = j;
    ret += cur[best];
    taken[best] = true;
    {
      int lo = a[best];
      for (int j = best - 1; j >= 0; --j) {
        lo = min(lo, a[j]);
        cur[j] -= (long long)2 * lo;
      }
    }
    {
      int lo = a[best];
      for (int j = best + 1; j < n; ++j) {
        lo = min(lo, a[j]);
        cur[j] -= (long long)2 * lo;
      }
    }
  }
  return ret;
}
void run() {
  scanf("%d%d", &n, &ntake);
  for (int i = (0); i < (n); ++i) scanf("%d", &a[i]);
  printf("%lld\n", solve());
}
int main() {
  run();
  return 0;
}

### Prompt

Generate a cpp solution to 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;
std::mt19937 rnd(
    (int)std::chrono::steady_clock::now().time_since_epoch().count());
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
const int MAXN = 4000;
int n, ntake;
int a[MAXN];
bool taken[MAXN];
long long cur[MAXN];
long long solve() {
  for (int i = (0); i < (n); ++i) taken[i] = false;
  for (int i = (0); i < (n); ++i) cur[i] = (long long)(ntake - 1) * a[i];
  long long ret = 0;
  for (int i = (0); i < (ntake); ++i) {
    int best = -1;
    for (int j = (0); j < (n); ++j)
      if (!taken[j] && (best == -1 || cur[j] > cur[best])) best = j;
    ret += cur[best];
    taken[best] = true;
    {
      int lo = a[best];
      for (int j = best - 1; j >= 0; --j) {
        lo = min(lo, a[j]);
        cur[j] -= (long long)2 * lo;
      }
    }
    {
      int lo = a[best];
      for (int j = best + 1; j < n; ++j) {
        lo = min(lo, a[j]);
        cur[j] -= (long long)2 * lo;
      }
    }
  }
  return ret;
}
void run() {
  scanf("%d%d", &n, &ntake);
  for (int i = (0); i < (n); ++i) scanf("%d", &a[i]);
  printf("%lld\n", solve());
}
int main() {
  run();
  return 0;
}

```