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 BASE = 1e9 + 7;

long long f[100100], fact[100100];

long long power(long long x, long long y)
{
  if (!y)
    return 1;
  long long res = power(x, y >> 1);
  res = res * res % BASE;
  if (y & 1)
    res = res * x % BASE;
  return res;
}

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

  int n, x[100100];
  cin >> n;
  for (int i = 1; i <= n; i++)
    cin >> x[i];

  fact[0] = 1;
  for (int i = 1; i <= n; i++)
    fact[i] = fact[i - 1] * i % BASE;
  f[1] = 1;
  for (int i = 2; i <= n; i++)
    f[i] = (f[i - 1] * i + fact[i - 1]) % BASE;

  long long ans = 0;
  for (int i = 1; i < n; i++)
  {
    long long ways = f[i] * fact[n - 1] % BASE * power(fact[i], BASE - 2) % BASE;
    ans = (ans + ways * (x[i + 1] - x[i])) % BASE;
  }
  cout << ans << endl;
}
