//预计10分
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define LL long long
using namespace std;int n;int h[320],vis[320];LL ans;void dfs(int now,int a,LL scor)
{if(a==n){ans=max(ans,scor);return ;}for(int i=1;i<=n;i++){if(vis[i]) continue;vis[i]=true;dfs(i,a+1,scor+(h[now]-h[i])*(h[now]-h[i]));vis[i]=false;}
}int main()
{freopen("rand.txt","r",stdin);freopen("baoli.txt","w",stdout);scanf("%d",&n);for(int i=1;i<=n;i++) scanf("%d",&h[i]);dfs(0,0,0);cout<<ans<<endl;return 0;
}

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define N 605
using namespace std;int cnt,g,flag,n;int a[N][N],qx[N*N],qy[N*N],vis[N][N];int h,t;int dx[4]={0,1,0,-1},dy[4]={-1,0,1,0};char b[N][N];struct B
{int x,y;
}c[N*N];bool canGo(int x,int y,int flag)
{a[x][y]=(flag%2==0?1:-1);              //先放上要走的棋子  memset(vis,0,sizeof(vis));for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){if(vis[i][j]||a[i][j]==0) continue;                    //*****h=1;t=0;int js=0;           //扫描联通块 qx[++t]=i;qy[t]=j;while(h<=t){    int nowx=qx[h],nowy=qy[h];h++;vis[nowx][nowy]=true;for(int v=0;v<4;v++)                                       //*****
                {int nxtx=nowx+dx[v],nxty=nowy+dy[v]; //遍历四周     if(nxtx<=0||nxty<=0||nxtx>n||nxty>n||vis[nxtx][nxty]) continue;     //*****if(!a[nxtx][nxty]) js++;              //存在气 if(a[nxtx][nxty]==a[nowx][nowy])    {qx[++t]=nxtx;qy[t]=nxty;                            //*****
                    }}}//    cout<<"--------------"<<endl;if(!js)                   //扫描的联通块没有气
            {a[x][y]=0;            //尝试失败 return false;}}}return true;
}int main()
{scanf("%d",&n);for(int i=1;i<=n;i++){cin>>b[i];for(int j=1;j<=n;j++){if(b[i][j-1]=='.')             //若为空
            {a[i][j]=0;c[++cnt].x=i;c[cnt].y=j;     //标记空位
            }if(b[i][j-1]=='X') a[i][j]=-1;     //若为黑棋 -1 if(b[i][j-1]=='O') a[i][j]=1;      //若为白棋 1
        }}if(n==1){printf("-1 -1\n");return 0;}      //棋盘只有一个空位，先手没法走 while(1){flag++;g=0;                          //flag标记当前是黑棋还是白棋走 for(int i=1;i<=cnt;i++)             //找一个能走的空位
      {int xx,yy;xx=c[i].x;yy=c[i].y;if(a[xx][yy]) continue;if(canGo(xx,yy,flag)) {printf("%d %d\n",xx,yy);g=true;a[xx][yy]=(flag%2==0?1:-1);break;}}if(!g){printf("-1 -1\n");return 0;}}return 0;
}