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;
static const int64_t mod=1000000007;
int64_t f(int64_t x,int64_t y){
if(y==0)
  return 1;
  else if(y%2==0)
    return (f(x,y/2)*f(x,y/2))%mod;
  else
    return (x*((f(x,y/2)*f(x,y/2))%mod))%mod;
}
int main(){
  int N;cin>>N;
  vector<int64_t>X(N);
  for(int i=0;i<N;i++)
    cin>>X.at(i);
  vector<int64_t>d(N-1);
  for(int i=0;i<N-1;i++)
    d.at(i)=X.at(i+1)-X.at(i);
  int64_t a=1;
  for(int64_t i=1;i<=N-1;i++)
    a=(a*i)%mod;
  int64_t ans=0;
  int64_t b=a;
  for(int i=0;i<N-1;i++){
  ans=(ans+b*d.at(i))%mod;
    b=(b+a*f(i+2,mod-2))%mod;
  }cout<<ans<<endl;
   return 0;
}