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.Scanner;

public class Main {
  public static void main(String[] args) {
    Scanner scanner = new Scanner(System.in);

    final long M = 1_000_000_007;
    int n = scanner.nextInt();
    int m = scanner.nextInt();
    int[] x = new int[n];
    int[] y = new int[m];
    for (int i = 0; i < n; i++) x[i] = scanner.nextInt();
    for (int i = 0; i < m; i++) y[i] = scanner.nextInt();

    System.out.println(compute(x, M) * compute(y, M) % M);
  }

  private static long compute(int[] a, long mod) {
    long ans = 0;
    int n = a.length;
    for (int i = 1; i < n; i++) {
      long d = (long) i * (n - i) % mod;
      d *= ((long) a[i] - a[i - 1]) % mod;
      d %= mod;
      ans += d;
      ans %= mod;
    }
    return ans;
  }
}