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 <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <cstdio>
#include <fstream>
#include <map>
#include <vector>
//#include <unordered_map>
//#include <unordered_set>
#include <stack>
#include <set>
#include <queue>
#include <list>

using namespace std;
const long long mod=1e9+7;
int x[100005];
int y[100005];
int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    long long lx=0,k=0,su=0;
    for(int i=0;i<n;++i)
        scanf("%d",&x[i]);
    for(int i=0;i<m;++i)
        scanf("%d",&y[i]);
    sort(x,x+n);
    sort(y,y+m);
    for(int i=1;i<n;++i)
    {
        k+=(x[i]-x[i-1])*i;
        k=(k%mod+mod)%mod;
        lx+=k;
        lx=(lx%mod+mod)%mod;
    }
    k=0;
    for(int i=1;i<m;++i)
    {
        k+=(y[i]-y[i-1])*i;
        k=(k%mod+mod)%mod;
        su+=(lx*k)%mod;
        su=(su%mod+mod)%mod;
    }
    cout<<su<<endl;
    return 0;
}