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<cstdio>
#include<cstdlib>
#include<cassert>
#include<cmath>
#include<queue>
#include<cstring>
#include<climits>
#include<sstream>
#include<deque>
#include<vector>
#include<algorithm>
#include<set>
#include<map>
#include<bitset>

#define REP(i,s,n) for(int i=s;i<n;++i)
#define rep(i,n) REP(i,0,n)

using namespace std;

typedef long long ll;

const ll mod = 1000000007LL;

ll n,m,x[100010], y[100100];

void compute() {
  ll sumX = 0;
  REP(i,1,n+1) {
    ll v1 = ( x[i] * (ll)(i-1) ) % mod;
    ll v2 = ( x[i] * (ll)(n-i) ) % mod;
    ll v  = ( v1 - v2 ) % mod;
    ( sumX += v ) %= mod;
  }
  ll sumY = 0;
  REP(i,1,m+1) {
    ll v1 = ( y[i] * (ll)(i-1) ) % mod;
    ll v2 = ( y[i] * (ll)(m-i) ) % mod;
    ll v  = ( v1 - v2 ) % mod;
    ( sumY += v ) %= mod;
  }
  cout << ( sumX * sumY ) % mod << endl;
}

int main() {
  
  cin >> n >> m;
  REP(i,1,n+1) cin >> x[i];
  REP(i,1,m+1) cin >> y[i];
  compute();
  return 0;
}