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;
int main(){
  int n;
  cin>>n;
  vector<int> x(n);
  for(int i=0;i<n;i++)
    cin>>x.at(i);
  int p=1000000007;
  int64_t f=1;
  for(int i=1;i<n;i++)
    f=f*i%p;
  vector<int64_t> in(n,1);
  int q=p-2;
  vector<bool> b;
  while(q>0){
    b.push_back(q%2==1);
    q/=2;
  }
  reverse(b.begin(),b.end());
  for(int i=2;i<n;i++){
    for(int d=0;d<b.size();d++){
      in.at(i)=in.at(i)*in.at(i)%p;
      if(b.at(d))
        in.at(i)=in.at(i)*i%p;
    }
  }
  int64_t s=0;
  int64_t c=0;
  for(int i=1;i<n;i++){
    c=(c+f*in.at(i))%p;
    s=(s+c*(x.at(i)-x.at(i-1)))%p;
  }
  cout<<s<<endl;
}