#include <bits/stdc++.h>
using namespace std;
template <typename T>
bool chkmax(T &x, T y) {
  return x < y ? x = y, true : false;
}
template <typename T>
bool chkmin(T &x, T y) {
  return x > y ? x = y, true : false;
}
int readint() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, m, rt;
long long a[4005], rmq[4005][20], ch[4005][2], lg2[4005], siz[4005];
long long d[4005][4005];
int getmin(int l, int r) {
  int k = lg2[r - l + 1];
  return a[rmq[l][k]] < a[rmq[r - (1 << k) + 1][k]] ? rmq[l][k]
                                                    : rmq[r - (1 << k) + 1][k];
}
int build(int l, int r) {
  int mid = getmin(l, r);
  if (l != mid) ch[mid][0] = build(l, mid - 1);
  if (r != mid) ch[mid][1] = build(mid + 1, r);
  return mid;
}
void dfs(int u) {
  if (!u) return;
  siz[u] = 1;
  dfs(ch[u][0]), dfs(ch[u][1]);
  siz[u] += siz[ch[u][0]] + siz[ch[u][1]];
  for (int i = 0; i <= siz[ch[u][0]]; i++) {
    for (int j = 0; j <= siz[ch[u][1]]; j++) {
      chkmax(d[u][i + j], d[ch[u][0]][i] + d[ch[u][1]][j] - 2ll * i * j * a[u]);
      chkmax(d[u][i + j + 1], d[ch[u][0]][i] + d[ch[u][1]][j] -
                                  2ll * i * j * a[u] + 1ll * (m - 1) * a[u] -
                                  2ll * (i + j) * a[u]);
    }
  }
}
int main() {
  n = readint();
  m = readint();
  for (int i = 1; i <= n; i++) a[i] = readint();
  for (int i = 2; i <= n; i++) lg2[i] = lg2[i >> 1] + 1;
  for (int i = 1; i <= n; i++) rmq[i][0] = i;
  for (int i = 1; (1 << i) <= n; i++)
    for (int j = 1; j + (1 << i) - 1 <= n; j++)
      rmq[j][i] = a[rmq[j][i - 1]] < a[rmq[j + (1 << (i - 1))][i - 1]]
                      ? rmq[j][i - 1]
                      : rmq[j + (1 << (i - 1))][i - 1];
  rt = build(1, n);
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= n; j++) d[i][j] = -(1ll << 60);
  d[0][0] = 0;
  dfs(rt);
  printf("%lld\n", d[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 namespace std;
template <typename T>
bool chkmax(T &x, T y) {
  return x < y ? x = y, true : false;
}
template <typename T>
bool chkmin(T &x, T y) {
  return x > y ? x = y, true : false;
}
int readint() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, m, rt;
long long a[4005], rmq[4005][20], ch[4005][2], lg2[4005], siz[4005];
long long d[4005][4005];
int getmin(int l, int r) {
  int k = lg2[r - l + 1];
  return a[rmq[l][k]] < a[rmq[r - (1 << k) + 1][k]] ? rmq[l][k]
                                                    : rmq[r - (1 << k) + 1][k];
}
int build(int l, int r) {
  int mid = getmin(l, r);
  if (l != mid) ch[mid][0] = build(l, mid - 1);
  if (r != mid) ch[mid][1] = build(mid + 1, r);
  return mid;
}
void dfs(int u) {
  if (!u) return;
  siz[u] = 1;
  dfs(ch[u][0]), dfs(ch[u][1]);
  siz[u] += siz[ch[u][0]] + siz[ch[u][1]];
  for (int i = 0; i <= siz[ch[u][0]]; i++) {
    for (int j = 0; j <= siz[ch[u][1]]; j++) {
      chkmax(d[u][i + j], d[ch[u][0]][i] + d[ch[u][1]][j] - 2ll * i * j * a[u]);
      chkmax(d[u][i + j + 1], d[ch[u][0]][i] + d[ch[u][1]][j] -
                                  2ll * i * j * a[u] + 1ll * (m - 1) * a[u] -
                                  2ll * (i + j) * a[u]);
    }
  }
}
int main() {
  n = readint();
  m = readint();
  for (int i = 1; i <= n; i++) a[i] = readint();
  for (int i = 2; i <= n; i++) lg2[i] = lg2[i >> 1] + 1;
  for (int i = 1; i <= n; i++) rmq[i][0] = i;
  for (int i = 1; (1 << i) <= n; i++)
    for (int j = 1; j + (1 << i) - 1 <= n; j++)
      rmq[j][i] = a[rmq[j][i - 1]] < a[rmq[j + (1 << (i - 1))][i - 1]]
                      ? rmq[j][i - 1]
                      : rmq[j + (1 << (i - 1))][i - 1];
  rt = build(1, n);
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= n; j++) d[i][j] = -(1ll << 60);
  d[0][0] = 0;
  dfs(rt);
  printf("%lld\n", d[rt][m]);
  return 0;
}

```