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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> P;
typedef pair<ll, ll> Pll;
#define debug(var)  do{std::cout << #var << " : ";view(var);}while(0)
template<typename T> void view(T e){std::cout << e << std::endl;}
template<typename T> void view(const std::vector<T>& v){for(const auto& e : v){ std::cout << e << " "; } std::cout << std::endl;}
template<typename T> void view(const std::vector<std::vector<T> >& vv){ for(const auto& v : vv){ view(v); } }
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

const int MOD = 1000000007;
const int INF = 1e9;

const int mod = 1000000007;
const int inf = 1e9;
#define PI acos(-1);
int dx[4] = {1,-1,0,0};
int dy[4] = {0,0,1,-1};
int ddx[8] = {1,1,1,-1,-1,-1,0,0};
int ddy[8] = {0,1,-1,0,1,-1,1,-1};

const ll MAX = 200000;
ll inv[MAX], fac[MAX], finv[MAX];

void settable() {
   fac[0] = fac[1] = 1;
   finv[0] = finv[1] = 1;
   inv[1] = 1;
   for (int i = 2; i < MAX ; i++){
       fac[i] = fac[i - 1] * i % MOD;
       inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
       finv[i] = finv[i - 1] * inv[i] % MOD;
   }
}
ll choose(ll n, ll k){
   if (n < k) return 0;
   if (n < 0 || k < 0) return 0;
   return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

ll modpow(ll n, ll k){
   ll rtn = 1;
   while(k > 0){
       if(k&1) rtn = rtn * n % MOD;
       n = n * n % MOD;
       k = k >> 1;
   }
   return rtn;
}

int main(){
    settable();
    int n, m;
    cin >> n >> m;
    vector<ll> x(n), y(m);
    for(int i = 0; i < n; i++) cin >> x[i];
    for(int i = 0; i < m; i++) cin >> y[i];
    ll sumx, sumy;
    sumx = sumy = 0;
    for(int i = 0; i < n; i++) {
        sumx += i * x[i] - (n-1-i) * x[i];
        sumx %= MOD;
    }
    for(int i = 0; i < m; i++) {
        sumy += i * y[i] - (m-1-i) * y[i];
    }
    ll ans = sumx * sumy % MOD;
    cout << ans << endl;
}