Subarray Cuts

You are given an array of length n and a number k. Let’s pick k non-overlapping non-empty subarrays of the initial array. Let s _i be the sum of the i-th subarray in order from left to right. Compute the maximum value of the following expression:|s₁ - s₂| + |s₂ - s₃| + … + |s_k - 1 - s_k|

Here subarray is a contiguous part of an array.

Input

The first line of input contains two integers n and k. The second line contains n integers — the elements of the array. The absolute values of elements do not exceed 10⁴.

The problem consists of two subproblems. The subproblems have different constraints on the input. You will get some score for the correct submission of the subproblem. The description of the subproblems follows.

• In subproblem E1 (9 points), constraints 2 ≤ n ≤ 400, 2 ≤ k ≤ min(n, 50) will hold.
• In subproblem E2 (12 points), constraints 2 ≤ n ≤ 30000, 2 ≤ k ≤ min(n, 200) will hold.

Output

Output a single integer — the maximum possible value.

Examples

input

5 3
5 2 4 3 1

output

12

input

4 2
7 4 3 7

output

8

Note

Consider the first sample test. The optimal solution is obtained if the first subarray contains the first element only, the second subarray spans the next three elements and the last subarray contains the last element only. The sums of these subarrays are 5, 9 and 1, correspondingly.

Consider the second sample test. In the optimal solution, the first subarray consists of the first two elements and the second subarray consists of the third element only. Note that the last element does not belong to any subarray in this solution.

Solution:

#include <cstring>
#include <vector>
#include  	<list>
#include

<map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <memory.h>
#include <cassert>

using namespace std;

const int inf = (int)1e9;
const int N = 30010;
const int K = 210;

int f[N][K][2][3][3];
int a[N];

void check(int &a, int b) {
if (b > a) a = b;
}

int main() {
int n, k;
scanf("%d %d", &n, &k);
for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
  }
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j <= k; j++) {
      for (int in = 0; in < 2; in++) {
        for (int c1 = 0; c1 < 3; c1++) {
          for (int c2 = 0; c2 < 3; c2++) {
            f[i][j][in][c1][c2] = -inf;
          }
        }
      }
    }
  }
  f[0][0][0][1][1] = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= k; j++) {
      for (int in = 1; in >= 0; in--) {
for (int c1 = 0; c1 < 3; c1++) {
          for (int c2 = 0; c2 < 3; c2++) {
            int ft = f[i][j][in][c1][c2];
            if (ft == -inf) {
              continue;
            }
            if (!in) {
              check(f[i + 1][j][0][c1][c2], ft);
              if (j < k) {
                for (int nc2 = 0; nc2 < 3; nc2++) {
                  if ((nc2 == 1) != (j + 1 == k)) {
                    continue;
                  }
                  check(f[i + 1][j + 1][1][c1][nc2], ft + a[i] * ((c1 - 1) + (nc2 - 1)));
                }
              }
            } else {
              check(f[i + 1][j][1][c1][c2], ft + a[i] * ((c1 - 1) + (c2 - 1)));
              check(f[i][j][0][2 - c2][1], ft);
            }
          }
        }
      }
    }
  }
  printf("%d\n", max(f[n][k][0][1][1], max(f[n][k][1][2][1], f[n][k][1][0][1])));
  return 0;
}