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].
#include <bits/stdc++.h>
using namespace std;
#define erorp(x) cout<<#x<<"={"<<(x.F)<<" , "<<x.S<<"}"<<endl
#define print(v,r) f(i,0,r) cout<<v[i]<<" "; cout<<endl
#define is_bit(x,y) (x%(1<<(y+1))>=(1<<y))
#define eror(x) cout<<#x<<'='<<(x)<<endl
#define f_(i,a,b) for(int i=a;i>=b;i--)
#define f(i,a,b) for(int i=a;i<b;i++)
#define nb(x) builtin_popcount(x)
#define maxm(a,b) a=max(a,b)
#define minm(a,b) a=min(a,b)
#define Add(x,y) x=(x+y)%mod
#define lst(x) x[x.size()-1]
#define sz(x) int(x.size())
#define mp make_pair
#define ll long long
#define pb push_back
#define S second
#define F first
#define int ll

const int N=5005,Max=(1ll<<31);

int n,m,c,a[N];
ll dp[N][N];

void solve(int l,int r,int x){
	cout<<l<<" "<<r<<endl;
	int s=++c,t=++c;
	if(r<=l) return ;
	if(l+1==r){
		dp[x][1]=a[l]*m-a[l];
		return ;
	}
	pair<int,int> mn=mp(Max,Max);
	f(i,l,r) minm(mn,mp(a[i],i));
	if(mn.S==l){
		solve(l,mn.S+1,s);
		solve(mn.S+1,r,t);
	}
	else{
		solve(l,mn.S,s);
		solve(mn.S,r,t);
	}
	f(i,0,max(1ll,mn.S-l)+1){
		f(j,0,min(r-l-1,r-mn.S)+1){
			maxm(dp[x][i+j],dp[s][i]+dp[t][j]-2*i*j*mn.F);
		}
	}
}

main(){
	ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
	cin>>n>>m;
	f(i,1,n+1) cin>>a[i];
	solve(1,n+1,0);
	cout<<dp[0][m];
}
