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)

#pragma region head
#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define fi first
#define se second
#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define repp(i, a, b) for (int i = a; i <= (b); ++i)
#define repr(i, a, b) for (int i = a; i >= (b); --i)
#define bit(n) (1LL << (n))
#define len(x) ((ll)(x).size())
#define debug(var) cout << "[" << #var << "]\n" << var << endl
#define int long long
typedef long long ll;
#define double long double
typedef double ld;
const int INF = 1001001001001001001ll;
const ll LINF = 1001001001001001001ll;
const int MOD = 1000000007;
const double EPS = 1e-9;
const double PI = acos(-1.0);

template <typename T>
ostream& operator<<(ostream& s, const vector<T>& v) {
  int len = v.size();
  for (int i = 0; i < len; ++i) {
    s << v[i];
    if (i < len - 1)
      s << ' ';
  }
  return s;
}

template <typename T>
ostream& operator<<(ostream& s, const vector<vector<T>>& vv) {
  int len = vv.size();
  for (int i = 0; i < len; ++i) {
    s << vv[i];
    if (i != len - 1)
      s << '\n';
  }
  return s;
}

template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}

template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}

__attribute__((constructor)) void initial() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(15);
}
#pragma endregion

#pragma region mod
struct mint {
  ll x;
  mint(ll x = 0) : x((x % MOD + MOD) % MOD) {
  }
  mint& operator+=(const mint a) {
    if ((x += a.x) >= MOD)
      x -= MOD;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += MOD - a.x) >= MOD)
      x -= MOD;
    return *this;
  }
  mint& operator*=(const mint a) {
    (x *= a.x) %= MOD;
    return *this;
  }
  mint operator+(const mint a) const {
    mint res(*this);
    return res += a;
  }
  mint operator-(const mint a) const {
    mint res(*this);
    return res -= a;
  }
  mint operator*(const mint a) const {
    mint res(*this);
    return res *= a;
  }

  /* 変数名.pow(n) */
  mint pow(ll t) const {
    if (!t)
      return 1;
    mint a = pow(t >> 1);
    a *= a;
    if (t & 1)
      a *= *this;
    return a;
  }

  // 以降はMODが素数のみ
  mint inv() const {
    return pow(MOD - 2);
  }
  mint& operator/=(const mint a) {
    return (*this) *= a.inv();
  }
  mint operator/(const mint a) const {
    mint res(*this);
    return res /= a;
  }

  // 入出力
  friend ostream& operator<<(ostream& s, const mint a) {
    s << a.x;
    return s;
  }
  friend istream& operator>>(istream& s, mint& a) {
    s >> a.x;
    return s;
  }
};

/* a^n (MOD) の計算 O(log n) */
ll modpow(ll a, ll n) {
  ll res = 1;
  while (n > 0) {
    if (n & 1)
      res = res * a % MOD;
    a = a * a % MOD;
    n >>= 1;
  }
  return res;
}

// a^{-1} mod を計算する
ll modinv(ll a) {
  return modpow(a, MOD - 2);
}
#pragma endregion

signed main() {
  int n, m;
  cin >> n >> m;
  vector<int> x(n);
  rep(i, n) {
    cin >> x[i];
  }
  vector<int> y(m);
  rep(i, m) {
    cin >> y[i];
  }

  mint ans = 1;

  vector<int> xx(n + 1);
  rep(i, n) {
    xx[i + 1] = xx[i] + x[i];
  }
  int tempx = 0;
  repp(j, 1, n - 1) {
    tempx += x[j] * j - xx[j];
  }
  ans *= tempx;

  vector<int> yy(m + 1);
  rep(i, m) {
    yy[i + 1] = yy[i] + y[i];
  }
  int tempy = 0;
  repp(j, 1, m - 1) {
    tempy += y[j] * j - yy[j];
  }
  ans *= tempy;

  cout << ans << endl;
}