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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;

#define DIV 1000000007

long long n, m;
long long x[100005];
long long y[100005];

int main(){
	cin >> n >> m;
	for(long long i = 0; i < n; i++){
		cin >> x[i];
	}
	for(long long i = 0; i < m; i++){
		cin >> y[i];
	}


	long long xsum = 0;
	for(long long i = 0; i < n; i++){
		//plus
		xsum += x[i] * max((long long)0, i);
		//xsum = (xsum + DIV)%DIV;
		//minus
		xsum -= x[i] * max((long long)0, n - 1 - i);
		//xsum = (xsum + DIV)%DIV;
	}

	long long ysum = 0;
	for(long long i = 0; i < m; i++){
		//plus
		ysum += y[i] * max((long long)0, i);
		//ysum = (ysum + DIV)%DIV;
		//minus
		ysum -= y[i] * max((long long)0, m - 1 - i);
		//ysum = (ysum + DIV)%DIV;
	}
	cout << (xsum * ysum)%DIV << endl;
}