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>
#include <vector>
using namespace std;

#define readv(v) for(auto& a : v) cin >> a;

using ll = long long int;
using vi = vector<ll>;
using vvi = vector<vi>;

const ll MOD = 1e9+7;

ll power(ll b, ll e){
  if (e == 1) return b;
  if (e % 2 == 0) return power((b*b)%MOD, e/2);
  return (b * power(b, e-1)) % MOD;
}

ll inverse(int i) {
  return power(i, MOD-2);
}

int main(){
  int N; cin >> N;
  vi x(N); readv(x);

  ll fact = 1; // (N-1)!
  for (int i = 2; i < N; i++) fact = (fact * i) % MOD;

  vi cum_sum(N, 0); // cum_sum[i] = /sum_{j=1}^{i} 1/j * (N-1)!
  for (int i = 1; i < N; i++) cum_sum[i] = (cum_sum[i-1] + fact * inverse(i)) % MOD;

  ll total = 0;
  for (int i = 0; i < N-1; i++) total = (total + ((x[i+1] - x[i]) * cum_sum[i+1]) % MOD) % MOD;

  cout << total << endl;
}
