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<long long>A(N-1,0), B(N-1,0), C(N,0);
  long long a=1, z=1e9+7;
  for(int i=1; i<N; i++){
    a*=i, a%=z, A[i-1]=a;
  }
  a=1;
  for(int i=N-1; i>0; i--){
    a*=i, a%=z, B[(N-1)-i]=a;
  }
  for(int i=0; i<N; i++){
    cin>>a;
    C[i]=a;
  }
  long long ans=0;
  for(int i=0; i<N-1; i++){
    if(i==0) ans+=((C[N-1]-C[i])*B[N-3-i])%z, ans%=z;
    else if(i==N-2) ans+=((C[N-1]-C[i])*A[i-1])%z, ans%=z;
    else ans+=((C[N-1]-C[i])*((A[i-1]*B[N-3-i])%z))%z, ans%=z;
  }
  cout<<ans<<endl;
}
