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: Accepted)

#include<iostream>
#include<cstdint>
using namespace std;

const int MOD = 1000000007; // 今回使う余り
int main() {
	int n, m, x, y;
	cin >> n >> m;
	
	int64_t xsum = 0, ysum = 0;
	for (int i = 0;i < n;i ++) {
		cin >> x;
		xsum = (xsum + (int64_t)(2 * i + 1 - n) * x) % MOD;
	}
	for (int i = 0;i < m;i ++) {
		cin >> y;
		ysum = (ysum + (int64_t)(2 * i + 1 - m) * y) % MOD;
	}
	cout << xsum * ysum % MOD;
	return 0;
}
/*
31536000のコメント解説欄
ここテンプレで用意してるから、A問題とかだとこの先空欄の危険あり
また、コンテスト後に https://31536000.hatenablog.com/ で解説していると思うので、良かったら読んでねー

単に式変形すると解ける
xy独立なのでまずはxについて説明するよ
まず、x方向ではΣ[i=1, n] Σ[j=i + 1, n] (xj - xi)が答え
これだと2重ループだけど、あるkを用いてΣ[k = 1, n] ((k - 1)x_k - (n - k)x_k) = Σ[k = 1, n] ((2k - n - 1)x_k)に等しいと分かるとO(n)
y方向も同様にやって終わり
*/