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 <stdio.h>
#include <string>
#include <cstring>
#include <stdlib.h>
#include <math.h>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <iterator>
#pragma warning(disable:4996)
 
typedef long long ll;
#define MIN(a, b) ((a)>(b)? (b): (a))
#define MAX(a, b) ((a)<(b)? (b): (a))
#define LINF 9223372036854775807
#define INF 2147483647
#define MOD 1000000007
using namespace std;


int main(int argc, char* argv[])
{
    long n, m;
    scanf("%ld%ld", &n, &m);

    vector<long> x(n);
    vector<long> y(m);
    long i;
    for(i=0; i<n; i++) {
        scanf("%ld", &x[i]);
    }
    for(i=0; i<m; i++) {
        scanf("%ld", &y[i]);
    }
    sort(x.begin(), x.end());
    sort(y.begin(), y.end());

#if 0
    ll sumx=0;
    long j;
    for(i=0; i<n-1; i++) {
        for(j=i+1; j<n; j++) {
            sumx += (x[j]-x[i]);
            sumx %= MOD;
        }
    }

    ll sumy=0;
    for(i=0; i<m-1; i++) {
        for(j=i+1; j<m; j++) {
            sumy += (y[j]-y[i]);
            sumy %= MOD;
        }
    }

#else
    ll sumx=0;
    for(i=0; i<n; i++) {
        sumx = (sumx + (i-(n-1-i))*(ll)x[i]);
        if(sumx>0) sumx = sumx %MOD;
        else       sumx = (MOD - (-sumx) %MOD)%MOD;
    }

    ll sumy=0;
    for(i=0; i<m; i++) {
        sumy = (sumy + (i-(m-1-i))*(ll)y[i]);
        if(sumy>0) sumy = sumy %MOD;
        else       sumy = (MOD - (-sumy) %MOD)%MOD;
    }
#endif

    printf("%lld\n", sumx*sumy%MOD);

    return 0;
}