#include <bits/stdc++.h>
using namespace std;
using ll = int64_t;
#define rep(i,n) for(int i = 0; i < ((int)(n)); i++)   // 0-indexed昇順

int main() {
    ll N, M; cin >> N >> M;
    vector<ll> X(N), Y(M);
    rep(i,N) cin >> X.at(i);
    rep(i,M) cin >> Y.at(i);
    ll XA = 0, YA = 0;
    ll MOD = (ll)pow(10,9) + 7;
    for (ll i = 0; i < N; i++) XA += (i * X[i] - (N - 1 - i) * X[i]), XA %= MOD;
    for (ll i = 0; i < M; i++) YA += (i * Y[i] - (M - 1 - i) * Y[i]), YA %= MOD;
    cout << XA * YA % MOD << endl;
}

### Prompt

Develop a solution in cpp to the problem described below:

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.

Examples

Input

3 3
1 3 4
1 3 6


Output

60


Input

6 5
-790013317 -192321079 95834122 418379342 586260100 802780784
-253230108 193944314 363756450 712662868 735867677


Output

835067060

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = int64_t;
#define rep(i,n) for(int i = 0; i < ((int)(n)); i++)   // 0-indexed昇順

int main() {
    ll N, M; cin >> N >> M;
    vector<ll> X(N), Y(M);
    rep(i,N) cin >> X.at(i);
    rep(i,M) cin >> Y.at(i);
    ll XA = 0, YA = 0;
    ll MOD = (ll)pow(10,9) + 7;
    for (ll i = 0; i < N; i++) XA += (i * X[i] - (N - 1 - i) * X[i]), XA %= MOD;
    for (ll i = 0; i < M; i++) YA += (i * Y[i] - (M - 1 - i) * Y[i]), YA %= MOD;
    cout << XA * YA % MOD << endl;
}

```