import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    static long largestValue(int[] A) {
        // Return the largest value of any of A's nonempty subarrays.
        int max_so_far = -1000000, max_ending_here = 0,end=0,start1=0,start=0,x=A.length;
 
    for (int i = 0; i < x; i++)
    {
        max_ending_here = max_ending_here + A[i];
       /* while(max_so_far-A[start]>max_so_far+A[i])
        {
            start++;
        }*/
        
        if (max_so_far < max_ending_here)
        {
            max_so_far = max_ending_here;
            end=i;
        }
 
        if (max_ending_here < 0)
        {
            max_ending_here = 0;
            
        }
    }
     
        long sum1=0;
        for (int i = 0; i <= end; i++)
        {
            sum1+=A[i];
        }
   // System.out.println(start+" "+end+" "+max_so_far+" "+sum1);
        int f=0;
        if(sum1==max_so_far)
            f=1;
        
        if(f==0)  
        for (int i = 0; i <=end; i++)
        {
            if(A[i]<0)
             sum1-=A[i];
            else
                sum1+=A[i];
            
   // System.out.println(sum1);
            if(sum1==max_so_far)
            {
                start=i+1;
                break;
            }
        }
        long sum=0;
        for(int i=start;i<=end;i++)
        {
            for(int j=i+1;j<=end;j++)
            {
                sum+=A[i]*A[j];
            }
        }
       return sum;
        
    }

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int[] A = new int[n];
        for(int A_i = 0; A_i < n; A_i++){
            A[A_i] = in.nextInt();
        }
        long result = largestValue(A);
        System.out.println(result);
        in.close();
    }
}