#include <bits/stdc++.h>
using i64 = long long;
const int mxn = 4005;
int n, m, arr[mxn];
int ch[mxn][2];
int rt;
int build(int l, int r) {
  if (l > r) return 0;
  int mid = 0;
  for (int i = l; i <= r; ++i) {
    if (mid == 0 || arr[i] < arr[mid]) {
      mid = i;
    }
  }
  ch[mid][0] = build(l, mid - 1);
  ch[mid][1] = build(mid + 1, r);
  return mid;
}
int siz[mxn];
i64 dp[mxn][mxn];
void dfs(int u) {
  if (!u) return;
  dfs(ch[u][0]);
  dfs(ch[u][1]);
  siz[u] = 1 + siz[ch[u][0]] + siz[ch[u][1]];
  memset(dp[u], -0x3f, sizeof(dp[u]));
  for (int a = 0; a < (int)(std::min(siz[ch[u][0]], m) + 1); ++a) {
    for (int b = 0; b < (int)(std::min(siz[ch[u][1]], m) + 1); ++b) {
      dp[u][a + b] = std::max(dp[u][a + b], dp[ch[u][0]][a] + dp[ch[u][1]][b] -
                                                2LL * a * b * arr[u]);
    }
  }
  for (int a = std::min(m, siz[u] - 1); a >= 0; --a) {
    dp[u][a + 1] = std::max(
        dp[u][a + 1], dp[u][a] + 1LL * m * arr[u] - 2LL * a * arr[u] - arr[u]);
  }
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d", &arr[i]);
  rt = build(1, n);
  memset(dp[0], -0x3f, sizeof(dp[0]));
  dp[0][0] = 0;
  dfs(rt);
  printf("%lld\n", dp[rt][m]);
  return 0;
}

### Prompt

Please formulate 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 i64 = long long;
const int mxn = 4005;
int n, m, arr[mxn];
int ch[mxn][2];
int rt;
int build(int l, int r) {
  if (l > r) return 0;
  int mid = 0;
  for (int i = l; i <= r; ++i) {
    if (mid == 0 || arr[i] < arr[mid]) {
      mid = i;
    }
  }
  ch[mid][0] = build(l, mid - 1);
  ch[mid][1] = build(mid + 1, r);
  return mid;
}
int siz[mxn];
i64 dp[mxn][mxn];
void dfs(int u) {
  if (!u) return;
  dfs(ch[u][0]);
  dfs(ch[u][1]);
  siz[u] = 1 + siz[ch[u][0]] + siz[ch[u][1]];
  memset(dp[u], -0x3f, sizeof(dp[u]));
  for (int a = 0; a < (int)(std::min(siz[ch[u][0]], m) + 1); ++a) {
    for (int b = 0; b < (int)(std::min(siz[ch[u][1]], m) + 1); ++b) {
      dp[u][a + b] = std::max(dp[u][a + b], dp[ch[u][0]][a] + dp[ch[u][1]][b] -
                                                2LL * a * b * arr[u]);
    }
  }
  for (int a = std::min(m, siz[u] - 1); a >= 0; --a) {
    dp[u][a + 1] = std::max(
        dp[u][a + 1], dp[u][a] + 1LL * m * arr[u] - 2LL * a * arr[u] - arr[u]);
  }
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d", &arr[i]);
  rt = build(1, n);
  memset(dp[0], -0x3f, sizeof(dp[0]));
  dp[0][0] = 0;
  dfs(rt);
  printf("%lld\n", dp[rt][m]);
  return 0;
}

```