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;
const int N=2e5+20;
const ll mod=1e9+7;
//ÌâÒâ:¸ø³ömÌõÆ½ÐÐÓÚxÖáµÄÖ±Ïß:yi ,nÌõÆ½ÐÐÓÚyÖáµÄÖ±Ïßxi,n,m<=1e5,ÇóËÄÔª×é(i,j,k,l)×é³É²»Í¬µÄ¾ØÐÎÃæ»ýÖ®ºÍ,i,j<=n,k,l<=m; 
//¶Ôx,y sortºó ans=segma,segma(xj-xi)*(yl-yk)
//ans=segma(xj-xi)*segma(yl-yk)
//n<=1e5,ÈçºÎ¿ìËÙµÃµ½segma(xj-xi),¼ÆËãxkµÄ¹±Ï× xk¼ÓÁËk-1´Î ±»¼õÁËn-k´Î segma(xj-xi)=segma((k-1)xk-(n-k)xk)  
ll x[N],y[N];
int main()
{
	ll n,m;
	while(cin>>n>>m)
	{
		for(int i=1;i<=n;i++)
			scanf("%I64d",&x[i]);	
		for(int i=1;i<=m;i++)
			scanf("%I64d",&y[i]);
		ll ans=0,s1=0,t=0;
		sort(x+1,x+1+n);
		sort(y+1,y+1+m);
		for(ll i=1;i<=n;i++)
			s1=(s1+(i-1)*x[i]-(n-i)*x[i])%mod;
		for(ll i=1;i<=m;i++)
			t=(t+(i-1)*y[i]-(m-i)*y[i])%mod;
		ans=(s1*t)%mod;
		cout<<ans<<endl;
	}	
	return 0;
}