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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<stdint.h>
#include<tgmath.h>
#include<stdbool.h>
#include<sys/queue.h>
#include<time.h>
#define ll long long
#define min(a, b) ((a) > (b) ? (b) : (a))
#define max(a, b) ((a) > (b) ? (a) : (b))

const ll mod = 1000000007ll;
ll power(ll x, ll y){
    if(y == 0ll)return 1ll;
    ll res = 1ll, tmp = x;
    while(y){
        if(y&1){
            res *= tmp;
            res %= mod;
        }
        tmp *= tmp;
        tmp %= mod;
        y >>= 1;
    }
    return res;
}
ll COM(ll x, ll y){
    if(x < y) return 0;
    if(x < 0 || y < 0) return 0;
    ll res = 1;
    for(int i=0; i<y; i++){
        res *= x - i;
        res %= mod;
        res *= power(y-i, mod-2);
        res %= mod;
    }
    return res;
}

ll n, m, x[100100], y[100100], a=0, b=0;

ll hoge(ll nyaa, ll nyan){
    ll res = nyaa * (nyaa+1) % mod;
    res += (nyan - 2*nyaa) * nyaa;
    return res % mod;
}

main(){
    scanf("%lld%lld", &n, &m);
    for(int i=0; i<n; i++){
        scanf("%lld", &x[i]);
        if(i){
            a += (x[i]-x[i-1])*hoge(min(i, n-i), n-1);
            a %= mod;
        }
    }
    for(int i=0; i<m; i++){
        scanf("%lld", &y[i]);
        if(i){
            b += (y[i]-y[i-1])*hoge(min(i, m-i), m-1);
            b %= mod;
        }
    }
    printf("%lld\n", a * b % mod);
    return 0;
}