There are N slimes standing on a number line. The i-th slime from the left is at position x_i.

It is guaruanteed that 1 \leq x_1 < x_2 < \ldots < x_N \leq 10^{9}.

Niwango will perform N-1 operations. The i-th operation consists of the following procedures:

* Choose an integer k between 1 and N-i (inclusive) with equal probability.
* Move the k-th slime from the left, to the position of the neighboring slime to the right.
* Fuse the two slimes at the same position into one slime.



Find the total distance traveled by the slimes multiplied by (N-1)! (we can show that this value is an integer), modulo (10^{9}+7). If a slime is born by a fuse and that slime moves, we count it as just one slime.

Constraints

* 2 \leq N \leq 10^{5}
* 1 \leq x_1 < x_2 < \ldots < x_N \leq 10^{9}
* x_i is an integer.

Input

Input is given from Standard Input in the following format:


N
x_1 x_2 \ldots x_N


Output

Print the answer.

Examples

Input

3
1 2 3


Output

5


Input

12
161735902 211047202 430302156 450968417 628894325 707723857 731963982 822804784 880895728 923078537 971407775 982631932


Output

750927044
#include <bits/stdc++.h>
using namespace std;

const int MOD = 1000000007;

// mod. mでのaの逆元a{-1}を計算する
long long modinv(long long a, long long m)
{
  long long b = m, u = 1, v = 0;
  while (b)
  {
    long long t = a / b;
    a -= t * b;
    swap(a, b);
    u -= t * v;
    swap(u, v);
  }
  u %= m;
  if (u < 0)
    u += m;
  return u;
}

int main()
{
  long long N;
  cin >> N;
  long long MOD = 1000000007;
  vector<long long> x(N);
  for (int i = 0; i < N; i++)
  {
    cin >> x[i];
  }
  vector<long long> c(N);
  c[0] = 0;
  for (int i = 1; i < N; i++)
  {
    c[i] = c[i - 1] + 1 * modinv(i, MOD);
  }
  vector<long long> deltax(N);
  deltax[0] = 0;
  for (int i = 1; i < N; i++)
  {
    deltax[i] = x[i] - x[i - 1];
  }
  long long sum = 0;
  for (int i = 1; i < N; i++)
  {
    sum += c[i] * deltax[i];
    sum %= MOD;
  }
  for (int i = N - 1; i >= 1; i--)
  {
    sum *= i;
    sum %= MOD;
  }
  cout << sum << endl;
}