UPC - ### (combination regimen)

Learning is like rowing upstream

UPC-###

Title Description

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 10⁹+7.
That is, for every quadruple (i,j,k,l) satisfying 1≤i<j≤n and 1≤k<l≤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 10⁹+7.
Constraints
2≤n,m≤10⁵
−10⁹≤x₁<…<x_n≤10⁹
−10⁹≤y₁<…<y_m≤10⁹
x_i and y_i are integers.

translation

On a two-dimensional plane, on a parallel line is drawn parallel to the x-axis and m, n and parallel lines parallel to the y axis. A straight line parallel to the x axis, from the bottom up to the i-th parallel to the straight line x Y _i is represented, equally, from the left to the right by the i-th parallel to the straight line x, x _i is represented.
For each of these straight lines by a rectangular siege, they determined area, and outputs all of ten ^{. 9} total area after taking over and +7.
That is, for every four satisfying 1≤i <j≤n, 1≤k <l≤m the (i, j, k, l ) is determined for each X = X _I , X = X _J , Y = Y _K and Y = Y _L , and the area consisting of, and in 10 ^{. 9} take +7.
Constraints
-10 ^{. 9} ≤x _{. 1} <... <X _n- ≦ 10 ^{. 9}
-10 ^{. 9} ≤Y _{. 1} <... <Y _m ≦ 10 ^{. 9}
X _I and Y _I are integers

Entry

Input is given from Standard Input in the following format:
n m
x₁ x₂ … x_n
y₁ y₂ … y_m

translation

Input from the standard input, the following format:
nm
X _{. 1} X ₂ ... X _n-
Y _{. 1} Y ₂ ... Y _m

Export

Print the total area of the rectangles, modulo 10⁹+7.

translation

Output of ten ^{. 9} all area after modulo +7.

Sample Input

3 3
1 3 4
1 3 6

Sample Output

60

Hint

The figure below illustrates this input:

The following is the total of the rectangular area A 9, B, ..., as shown as 60. The figure below illustrates this input:

Analytical thinking

Summarize what is a formula
$\sum_{1≤a<b≤4}\sum_{1≤c<d≤4}\left ( b-a \right )\left ( d-c \right )\,$
If you still do not understand how one thing, took over the motion picture speak for themselves

and then with the delicious cheese mathematical formula can get open
$\sum_{1≤a<b≤4}\left ( b-a \right )*\sum_{1≤c<d≤4}\left ( d-c \right )\,$

After it is calculated according to this formula all the area
, but it is regular, painting at the map to find it

OK, attach codes
AC time to! !

#include<iostream>
#include<algorithm>
#include<string.h>
#include<map>
#include<string>
#include<math.h>
#include<stdio.h>
#pragma GCC optimize(2)
#define Swap(a,b)  a ^= b ^= a ^= b
using namespace std;
typedef long long ll;
const int MAXN=10010;
const ll long_inf=9223372036854775807;
const int int_inf=2147483647;
inline ll read() {
	ll c=getchar(),Nig=1,x=0;
	while(!isdigit(c)&&c!='-')c=getchar();
	if(c=='-')Nig=-1,c=getchar();
	while(isdigit(c))x=((x<<1)+(x<<3))+(c^'0'),c=getchar();
	return Nig*x;
}
ll N=1e9+7;
ll x[100005],y[100005];
int main() {
	ll a,b;
	cin>>a>>b;
	for(ll i=0; i<a; i++)
		x[i]=read();
	for(ll j=0; j<b; j++)
		y[j]=read();
	ll sumx=0,sumy=0;
	for(ll i=1; i<a; i++) {
		sumx+=(((((a-i)%N)*i)%N)*((x[i]-x[i-1])%N));
		sumx%=N;
	}
	for(ll i=1; i<b; i++) {
		sumy+=(((((b-i)%N)*i)%N)*((y[i]-y[i-1])%N));
		sumy%=N;
	}
	cout<<(sumx*sumy)%N<<endl;
}

If not (ni) * i traveled all the words, but will time out, so this place has played the role of optimization, to avoid double counting.

There is a road ground for the tracks, Boundless Learning bitter for the boat.

By- wheel month