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 MOD 1000000007
 
long long mod_inv(long long n)
{
    long long left=MOD-2;  // define MOD!
    long long num=n;
    long long ans=1;
    while(left!=0)
    {
        if(left%2==1)ans=(ans*num)%MOD;
        left/=2;
        num=(num*num)%MOD;
    }
    return ans;
}

int main()
{
  long long  n;cin>>n;
  vector<long long> x(n);
  long long i;
  for(i=0;i<n;i++)cin>>x.at(i);
  long long nk=1;
  for(i=1;i<=n-1;i++)
  {
    nk=(nk*i)%MOD;
  }
  vector<long long> rsw(n-1,nk);
  for(i=0;i<n-1;i++)
  {
    rsw.at(i)=(rsw.at(i)*mod_inv(i+1))%MOD;
  }
  for(i=1;i<n-1;i++)
  {
    rsw.at(i)=(rsw.at(i-1)+rsw.at(i))%MOD;
  }
  long long ans=0;
  for(i=0;i<n-1;i++)
  {
    ans=(ans+rsw.at(i)*(x.at(i+1)-x.at(i)))%MOD;
  }
  cout<<ans<<endl;
}