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 C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char const *argv[]) {
  int n, m, i, j, k, l;
  scanf("%d %d", &n , &m);


  long long *x;
  long long *y;
  x = malloc(sizeof(long long) * n);
  y = malloc(sizeof(long long) * m);
  for (i = 0; i < n; i++) {
    scanf("%lld", &x[i]);
  }
  for (i = 0; i < m; i++) {
    scanf("%lld", &y[i]);
  }

  long long sum = 0;

  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      for (k = 0; k < m; k++) {
        for (l = 0; l < m; l++) {
          if (x[i] >= 1 && x[i] <= 1000000000 && x[i] < x[j] && x[j] >= 1 && x[j] <= 1000000000) {
            if (y[k] >= 1 && y[k] <= 1000000000 && y[k] < y[l] && y[l] >= 1 && y[l] <= 1000000000) {
              //printf("%lld\n", (x[j] - x[i]) * (y[l] - y[k]));
              sum += (x[j] - x[i]) * (y[l] - y[k]);
            }
          }
        }
      }
    }
  }
  printf("%lld\n", sum % 1000000007);
  free(x);
  free(y);
  return 0;
}