T. E. Lawrence was a controversial figure during World War I. He was a British officer who served in the Arabian theater and led a group of Arab nationals in guerilla strikes against the Ottoman Empire. His primary targets were the railroads. A highly fictionalized version of his exploits was presented in the blockbuster movie, "Lawrence of Arabia".

You are to write a program to help Lawrence figure out how to best use his limited resources. You have some information from British Intelligence. First, the rail line is completely linear---there are no branches, no spurs. Next, British Intelligence has assigned a Strategic Importance to each depot---an integer from 1 to 100. A depot is of no use on its own, it only has value if it is connected to other depots. The Strategic Value of the entire railroad is calculated by adding up the products of the Strategic Values for every pair of depots that are connected, directly or indirectly, by the rail line. Consider this railroad: 


Its Strategic Value is 4*5 + 4*1 + 4*2 + 5*1 + 5*2 + 1*2 = 49.

Now, suppose that Lawrence only has enough resources for one attack. He cannot attack the depots themselves---they are too well defended. He must attack the rail line between depots, in the middle of the desert. Consider what would happen if Lawrence attacked this rail line right in the middle: 

The Strategic Value of the remaining railroad is 4*5 + 1*2 = 22. But, suppose Lawrence attacks between the 4 and 5 depots: 

The Strategic Value of the remaining railroad is 5*1 + 5*2 + 1*2 = 17. This is Lawrence‘s best option.

Given a description of a railroad and the number of attacks that Lawrence can perform, figure out the smallest Strategic Value that he can achieve for that railroad. 

4 1
4 5 1 2
4 2
4 5 1 2
0 0

17
2

/*分析：
假定dp[i][j]表示前i个数分成j段的最小值
cost[i]表示从1~i的数两两相乘的总和
sum[i]表示前i个数的和
则：
dp[i][j]=Min(dp[k][j-1]+cost[i]-cost[k]-sum[k]*(sum[i]-sum[k]))
=>dp[i][j]=dp[k][j-1]+cost[i]-cost[k]-sum[i]*sum[k]+sum[k]*sum[k]
由于有sum[i]*sum[k]这一项,所以不可能用单调队列维护
-cost[k]-sum[i]*sum[k]+sum[k]*sum[k]的最小值
所以我们要把sum[i]独立出来以便求维护单调队列是和i无关 
如今我们须要找出最优的k,
令k2<k时k时最优的,即前k个数k为最优的取值
所以满足：
dp[k][j-1]+cost[i]-cost[k]-sum[i]*sum[k]+sum[k]*sum[k]
<=dp[k2][j-1]+cost[i]-cost[k2]-sum[i]*sum[k2]+sum[k2]*sum[k2]
=>(dp[k][j-1]-cost[k]+sum[k]*sum[k]-(dp[k2][j-1]-cost[k2]+sum[k2]*sum[k2]))/(sum[k]-sum[k2])<=sum[i]
设:
y1=dp[k][j-1]-cost[k]+sum[k]*sum[k]
x1=sum[k]
y2=dp[k2][j-1]-cost[k2]+sum[k2]*sum[k2]
x2=sum[k2]
所以变成:
(y2-y1)/(x2-x1)
即两点之间的斜率!
对于这个斜率我们怎么来寻找关系维护?
假定k点之前k最优且k点在单调队列的首位置 
如今对于k+1点
首先对于队列中的点更新k+1之后的点能否比k,k+1..更优，即更新最长处(由于sum[k+1]添加了所以能够更新最长处)
然后对于点k+1与队列中最后一个点的斜率必须是整个队列中斜率最大的,即保持斜率单调添加
为什么？由于假如k+1与k的斜率比k与k-1的斜率更低,
则随着sum[k+x]添加,k+1会首先更优于k,所以不须要k点,仅仅须要比較k+1与k-1的点的优先值 
*/
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <queue>
#include <algorithm>
#include <map>
#include <cmath>
#include <iomanip>
#define INF 99999999
typedef long long LL;
using namespace std;

const int MAX=1000+10;
int n,m,index,head,tail;
int s[MAX],q[MAX];
int dp[MAX][2],cost[MAX],sum[MAX];//dp採用滚动数组 

int GetY(int k,int k2){
	return dp[k][index^1]-cost[k]+sum[k]*sum[k]-(dp[k2][index^1]-cost[k2]+sum[k2]*sum[k2]);
}

int GetX(int k,int k2){
	return sum[k]-sum[k2];
}

void DP(){
	index=0;
	memset(dp,0,sizeof dp);
	for(int i=1;i<=n;++i)dp[i][index]=cost[i];
	for(int j=1;j<=m;++j){//分成j段,j作为第一层循环才用滚动数组
		index=index^1; 
		head=tail=0;
		q[tail++]=0;
		for(int i=1;i<=n;++i){
			while(head+1<tail && GetY(q[head+1],q[head])<=GetX(q[head+1],q[head])*sum[i])++head;
			dp[i][index]=dp[q[head]][index^1]+cost[i]-cost[q[head]]-sum[i]*sum[q[head]]+sum[q[head]]*sum[q[head]];
			while(head+1<tail && GetY(i,q[tail-1])*GetX(q[tail-1],q[tail-2])<=GetY(q[tail-1],q[tail-2])*GetX(i,q[tail-1]))--tail;
			q[tail++]=i;
		}
	}
}

int main(){
	while(~scanf("%d%d",&n,&m),n+m){
		for(int i=1;i<=n;++i)scanf("%d",&s[i]);
		for(int i=1;i<=n;++i)sum[i]=sum[i-1]+s[i];
		memset(cost,0,sizeof cost);
		for(int i=1;i<=n;++i){
			for(int j=i+1;j<=n;++j)cost[j]+=s[i]*s[j];
		}
		for(int i=1;i<=n;++i)cost[i]+=cost[i-1];
		DP();
		printf("%d\n",dp[n][index]);
	}
	return 0;
}