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: Time Limit Exceeded)

#include "bits/stdc++.h"
#define MAXN 100009
#define INF 1000000007
#define mp(x,y) make_pair(x,y)
#define all(v) v.begin(),v.end()
#define pb(x) push_back(x)
#define wr cout<<"----------------"<<endl;
#define ppb() pop_back()
#define tr(ii,c) for(typeof((c).begin()) ii=(c).begin();ii!=(c).end();ii++)
#define ff first
#define ss second
#define my_little_dodge 46
using namespace std;

typedef long long ll;
typedef pair<int,int> PII;
template<class T>bool umin(T& a,T b){if(a>b){a=b;return 1;}return 0;}
template<class T>bool umax(T& a,T b){if(a<b){a=b;return 1;}return 0;}
int mod(ll x){
	while(x<0)
		x+=INF;
	return (x%INF);
}
int main(){
	//~ #ifndef ONLINE_JUDGE
    //~ freopen("file.in", "r", stdin);
	//~ #endif	
	int n,m;
	int A=0,B=0,sum=0;
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++){
		int x;
		scanf("%d",&x);
		A=mod(A+mod(mod((i-1)*1LL*x)-sum));
		sum=mod(sum+x);
	}sum=0;
	for(int i=1;i<=m;i++){
		int x;
		scanf("%d",&x);
		B=mod(B+mod(mod((i-1)*1LL*x)-sum));
		sum=mod(sum+x);
	}
	printf("%d\n",mod(A*1LL*B));
	return 0;
}