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;
typedef int64_t ll;
const int MOD=1e9+7;
int modinv(ll a)
{
  ll b=MOD,u=1,v=0;
  while(b)
  {
    ll t=a/b;
    a-=t*b;swap(a,b);
    u-=t*v;swap(u,v);
  }
  u%=MOD; 
  if(u<0)u+=MOD;
  return u;
}
int main()
{
  ll answer=0,N;
  cin>>N;
  vector<ll> s(N,0);
  for(int i=0;i<N;i++) cin>>s.at(i);
  for(int i=0;i<N-1;i++) answer+=((s.at(N-1)-s.at(i))*modinv(i+1))%MOD;
  for(int i=1;i<N;i++) answer=(answer*i)%MOD;
  cout<<answer<<endl;
}