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: Time Limit Exceeded)

import java.util.*;

public final class Main {

    public static void main(String[] args) {

        Scanner s = new Scanner(System.in);

        int n  = s.nextInt();

        int m = s.nextInt();

        long mod = 1000000000 + 7;

        long[] x = new long[n];

        for (int i = 0; i < n; i++) {

            x[i] = s.nextInt();
        }


        long[] y = new long[m];

        for (int i = 0; i < m; i++) {

            y[i] = s.nextInt();
        }

        long before = 0;

        long result = 0;

        for (int i = 1; i < x.length;i++) {

            long x2 = x[i];

            long x1 = x[i-1];

            long interval = x2 - x1;

            long sum = 0;
            long innerbefore = 0;

            for (int j = 1; j < y.length; j++) {

                long y2 = y[j];
                long y1 = y[j-1];

                long intv = y2 - y1;

                long local = interval * intv;

                long now = (local%mod * j)%mod + innerbefore;

                sum += now;

                innerbefore = now;

            }

            long k = before + (sum*i)%mod;

            result += k;

            before = k;
        }

        System.out.print(result%mod);
    }


}