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.io.*;
import java.util.*;
import java.lang.*;

public class Main {

    private static long MOD = 1000000007L;

    public static void main(String[] args) {
        Scanner fs = new Scanner(System.in);
        int n = fs.nextInt(), m = fs.nextInt();
        System.out.println((helper(fs, n) * helper(fs, m)) % MOD);
    }

    private static long helper(Scanner fs, int n) {
        long ans = 0L;
        Long[] xx = new Long[n];
        for (int i = 0; i < n; ++i) {
            xx[i] = fs.nextLong();
        }
        Arrays.sort(xx);
        long prev = xx[0];
        for (int i = 1; i < n; ++i) {
            long x = xx[i];
            long tmp = (x - prev) * i % MOD;
            tmp = tmp * (n - i) % MOD;
            ans = (ans + tmp) % MOD;
            prev = x;
        }
        return ans;
    }
}