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: Accepted)

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 sum_interval = 0;

        long before_inteval = 0;
        for (int i = 1; i< m; i++) {

            long interval = y[i] - y[i-1];

            long now = before_inteval + (interval*i)%mod;

            before_inteval = now;

            sum_interval += now;
        }

        sum_interval = sum_interval %mod;

        long result = 0;
        long before_sum = 0;
        for (int i = 1; i < n; i++) {

            long x1 = x[i];

            long x2 = x[i-1];

            long xint = x1-x2;

            long now = xint * sum_interval;

            long r = (now%mod*i)%mod + before_sum;

            before_sum = r;

            result += r;
        }

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


}