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 <iostream>
using namespace std;
#define MOD 1000000007
typedef long long ll;

int main() {
  int n;
  cin >> n;
  int x[n];
  for (int i = 0; i < n; i++) {
    cin >> x[i];
  }

  ll a[n], p = 1;
  a[1] = 1;
  for (int i = 2; i < n; i++) {
    p = p * (i-1) % MOD;
    a[i] = (a[i-1] * i + p) % MOD;
  }

  ll b[n];
  b[n-1] = 1;
  for (int i = n-2; i > 0; i--) {
    b[i] = b[i+1] * (i+1) % MOD;
  }

  ll ans = 0;
  for (int i = 1; i < n; i++) {
    ans += (x[i]-x[i-1]) * (a[i]*b[i]%MOD) % MOD;
    ans %= MOD;
  }
  cout << ans << endl;
}