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>
constexpr int MOD = 1000000007;
constexpr int MAX = 100010;
using ll = long long;
using namespace std;

ll inv(ll x) {
  ll res = 1;
  ll k = MOD - 2;
  ll y = x;
  while (k) {
    if (k & 1) res = (res * y) % MOD;
    y = (y * y) % MOD;
    k /= 2;
  }
  return res;
}

ll f[MAX];
ll h[MAX];
void init() {
  f[0] = 1;
  for (int i = 1; i < MAX; i++)
    f[i] = (f[i - 1] * i) % MOD;
  
  h[0] = 0;
  for (int i = 1; i < MAX; i++)
    h[i] = (h[i - 1] + inv(i)) % MOD;
}

int main() {
  init();
  int n;
  cin >> n;
  vector<ll> x(n);
  for (int i = 0; i < n; i++)
    cin >> x[i];
  
  vector<ll> d(n);
  d[0] = f[n - 1];
  for (int i = 1; i < n - 1; i++) {
    d[i] = f[n - 1] * h[i + 1];
    d[i] %= MOD;
  }
  
  ll total = 0;
  for (int i = 0; i < n - 1; i++) {
    total += d[i] * (x[i+1] - x[i]);
    total %= MOD;
  }
  
  cout << total << endl;
  
  return 0;
}
