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: Accepted)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using pii = pair<int, int>;
using piii = pair<int, pair<int, int>>;
using pll = pair<ll, ll>;
using pli = pair<ll, int>;
template <typename T>
using Graph = vector<vector<T>>;
const int MOD = 1e9 + 7;

ll sum(vector<ll> &vsum) {
    int siz = vsum.size();
    ll res = 0;
    for (int i = 1; i < siz; ++i) {
        ll v = vsum[i] - vsum[i - 1];
        res += (i - 1) * v - vsum[i - 1];
        res %= MOD;
    }
    return res;
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    int n, m;
    cin >> n >> m;
    vector<ll> xsum(n + 1), ysum(m + 1);
    for (int i = 0; i < n; ++i) {
        int x;
        cin >> x;
        xsum[i + 1] = (xsum[i] + x) % MOD;
    }
    for (int i = 0; i < m; ++i) {
        int y;
        cin >> y;
        ysum[i + 1] = (ysum[i] + y) % MOD;
    }

    ll ans = sum(xsum) * sum(ysum) % MOD;
    if (ans < 0)
        ans += MOD;
    cout << ans << endl;
    return 0;
}