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;
#define M 1000000007
long power(long x,long p){
  if(p==0)return 1;
  if(p%2)return x*power(x,p-1)%M;
  long a=power(x,p/2);return a*a%M;
}
int main(){
  long N;cin>>N;
  vector<long>inv(N+1);
  for(int i=1;i<=N;i++){inv[i]=power(i,M-2);}
  
  vector<long>X(N),C(N-1);for(int i=0;i<N;i++)cin>>X[i];
  for(int i=0;i<N-1;i++)C[i]=X[i+1]-X[i];
  long P=1;
  for(int i=1;i<=N-1;i++)P=(P*i)%M;
  long ans=0;
  vector<long>T(N);T[0]=P;
  for(int i=1;i<N-1;i++)T[i]=(T[i-1]+(P*inv[i+1])%M)%M;
  for(int j=0;j<N-1;j++){
    long tmp=T[j]%M;
    ans=(ans+(tmp*C[j])%M)%M;
    assert(P>=0);assert(tmp>=0);assert(ans>=0);
  }
  cout<<ans%M<<endl;
  
}


