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;

typedef long long int ll;

const int N = 1e5 + 5;
const ll mod = 1e9 + 7;

vector < ll > fact(N);

ll pwr (ll a, ll b) {
  ll res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}

ll inv (ll a) {
  return pwr(a, mod - 2);
}

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

  fact[0] = 1;
  for (int i = 1; i < N; i++) fact[i] = (i * fact[i - 1]) % mod;

  int n;
  cin >> n;
  vector < int > x(n + 1);
  vector < ll > p(n + 1);
  for (int i = 1; i <= n; i++) cin >> x[i];
  ll ans = 0;
  for (int i = 1; i <= n; i++) {
    p[i] = (p[i - 1] + inv(i)) % mod;
  }
  for (int i = 1; i < n; i++) {
    ans = (ans + (((x[i + 1] - x[i]) * p[i]) % mod)) % mod;
  }
  ans = (ans * fact[n - 1]) % mod;
  cout << ans << '\n';
}
