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 class Main {
    public static void main(String[] args) {
        final long DIV = 1000000007L;
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        long[] x = new long[n];
        long[] y = new long[m];
        for(int i=0; i<n; i++){
            x[i] = sc.nextLong();
        }
        for(int j=0; j<m; j++){
            y[j] = sc.nextLong();
        }
        sc.close();

        // long[] xmult = new long[n-1];
        long xmultsum = 0;
        for(int i=0; i<n-1; i++){
            // xmult[i] = (x[i+1] - x[i]) * (i+1) * (n-i);
            xmultsum += (x[i+1] - x[i]) * (i+1) * (n-i-1);
            xmultsum %= DIV;
        }
        long ans = 0;
        // long[] ymult = new long[m-1];
        for(int i=0; i<m-1; i++){
            // ymult[i] = (y[i+1] - y[i]) * (i+1) * (m-i);
            ans += xmultsum * (((y[i+1] - y[i]) * (i+1) * (m-i-1)) % DIV);
            ans %= DIV;
        }

        System.out.println(ans);
    }
}