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 int64_t mod = 1000000007;

int64_t power(int64_t n, int64_t k) {
  if (k == 0) return 1;
  int64_t half = power(n, k / 2), ans = half * half % mod;
  if (k % 2) {
    ans *= n;
    ans %= mod;
  }
  return ans;
}

int main() {
  int n;
  cin >> n;
  vector<int> x(n);
  for (int i = 0; i < n; i++) cin >> x.at(i);
  int64_t fact = 1, ans = 0;
  for (int i = 1; i < n; i++) fact = fact * i % mod;
  vector<int64_t> s(n);
  for (int i = 1; i < n; i++) {
    s.at(i) = (s.at(i - 1) + power(i, mod - 2)) % mod;
  }
  for (int i = n - 1; i > 0; i--) {
    int64_t dx = x.at(i) - x.at(i - 1);
    ans += dx * s.at(i) % mod;
    ans %= mod;
  }
  cout << fact * ans % mod << endl;
}
