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: Runtime Error)

#include <iostream>
#include<iomanip>
#include <set>
#include <vector>
#include <map>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <queue>
#include <stack>
#include <list>
using namespace std;

const int NN=1e5+10;
const int mod=1e9+7;

long long x[60],y[60];
long long xx[60],yy[60];
long long n,m;

int main()
{
    scanf("%lld%lld",&n,&m);
    for(int i=0;i<n;i++){
        scanf("%lld",&x[i]);
    }
    sort(x,x+n);
    for(int i=0;i<n-1;i++){
        xx[i]=x[i+1]-x[i];
    }

    for(int i=0;i<m;i++){
        scanf("%lld",&y[i]);
    }
    sort(y,y+m);
    for(int i=0;i<m-1;i++){
        yy[i]=y[i+1]-y[i];
    }
    long long nn=((n-1)/2);
    long long mm=((m-1)/2);
    long long sum1=0;
    long long sum2=0;
    for(int i=0;i<nn;i++){
        sum1=(sum1+xx[i]*(n-1+i))%mod;
        if((n-2-i)>0){
            sum1=((sum1+xx[n-2-i]*(n-1+i))%mod)%mod;
        }
    }
    if((n-1)%2!=0&&(n-1)!=1){
        sum1=(sum1+(xx[nn]*(n-1+nn)))%mod;
    }

    for(int i=0;i<mm;i++){
        sum2=(sum2+yy[i]*(m-1+i))%mod;
        if((m-2-i)>0){
            sum2=(sum2+yy[m-2-i]*(m-1-i))%mod;
        }
    }
    if((m-1)%2!=0&&(m-1)!=1){
        sum2=(sum2+(yy[mm]*(m-1+mm)))%mod;
    }
    //printf("%lld %lld\n",sum1,sum2);
    long long ans=(((sum1*sum2)%mod))%mod;
    printf("%lld\n",ans);
}