Problem:
Score : 500 points

Problem Statement

On a two-dimensional plane, there are m lines drawn parallel to the x axis, and n lines drawn parallel to the y axis.
Among the lines parallel to the x axis, the i-th from the bottom is represented by y = y_i.
Similarly, among the lines parallel to the y axis, the i-th from the left is represented by x = x_i.

For every rectangle that is formed by these lines, find its area, and print the total area modulo 10^9+7.

That is, for every quadruple (i,j,k,l) satisfying 1\leq i < j\leq n and 1\leq k < l\leq m, find the area of the rectangle formed by the lines x=x_i, x=x_j, y=y_k and y=y_l, and print the sum of these areas modulo 10^9+7.

Constraints

2 \leq n,m \leq 10^5

-10^9 \leq x_1 < ... < x_n \leq 10^9

-10^9 \leq y_1 < ... < y_m \leq 10^9

x_i and y_i are integers.

Input

Input is given from Standard Input in the following format:

n m
x_1 x_2 ... x_n
y_1 y_2 ... y_m

Output

Print the total area of the rectangles, modulo 10^9+7.

Sample Input 1

3 3
1 3 4
1 3 6

Sample Output 1

60

The following figure illustrates this input:

The total area of the nine rectangles A, B, ..., I shown in the following figure, is 60.

Sample Input 2

6 5
-790013317 -192321079 95834122 418379342 586260100 802780784
-253230108 193944314 363756450 712662868 735867677

Sample Output 2

835067060

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;

class Main
{
    public static void main (String[] args)
    {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        long[] a = new long[n];
        long[] x = new long[n];
        long[] b = new long[m];
        long[] y = new long[m];
        long MOD = (long) (Math.pow(10, 9) + 7);

        a[0] = sc.nextLong();
        for(int i = 1; i < n; i++){
            a[i] = sc.nextLong();
            x[i] = a[i] - a[i-1];
            x[i] %= MOD;
        }

        b[0] = sc.nextLong();
        for(int i = 1; i < m; i++){
            b[i] = sc.nextLong();
            y[i] = b[i] - b[i-1];
            y[i] %= MOD;
        }

        long ans = 0;

/*        long x_sum = 0;

        for(int i = 1; i < n; i++){
            long xx = (n - i) * ( i );    
            xx %= MOD;
            x[i] *= xx;
            x[i] %= MOD;
            x_sum += x[i];
            x_sum %= MOD;
        }

        long y_sum = 0;

        for(int i = 1; i < m; i++){
            long yy = (m - i) * ( i );    
            yy %= MOD;
            y[i] *= yy;
            y[i] %= MOD;
            y_sum += y[i];
            y_sum %= MOD;
        }*/

        long a_sum = 0;
        for(int i = 0; i < n; i++){
            a_sum += ( i * 2 - n + 1 ) * a[i];
            a_sum %= 2;
        }

        long b_sum = 0;
        for(int i = 0; i < m; i++){
            b_sum += ( i * 2 - m + 1 ) * b[i];
            b_sum %= 2;
        }

 //       ans = x_sum * y_sum;
        ans = a_sum * b_sum;
        ans %= MOD;
        System.out.println(ans);
    }
}