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 <math.h>
#include <vector>
using namespace std;

long long int modpow(long long a, long long n, long long mod) {
  long long int res = 1;
  while (n > 0) {
    if (n & 1) res = (res * a % mod)%1000000007;
    a = a * a % mod;
    n >>= 1;
  }
  return res%1000000007;
  }

long long modinv(long long a, long long mod) {
  return modpow(a, mod - 2, mod)%1000000007;
}

int main(){
  long long int N;
  vector<long long int> x;
  cin >> N;
  long long int N_ex=1;
  for(int i=0;i<N;i++){
    long long int tmp; cin >> tmp;
    x.push_back(tmp);
  }
  for(int i=1;i<=N-1;i++){
    N_ex=(N_ex*i)%1000000007;
  }
  long long int ans = 0;
  for(long long int i=1;i<=N-1;i++){
    ans =(ans + (N_ex*modinv(i, 1000000007) % 1000000007*(x[N-1]-x[i-1]) % 1000000007)) % 1000000007;
  }
  //if(ans<0){ans=ans+1000000007;}
  cout << ans;
  return 0;
}
