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 int long
const int mod = 1000000007;
int gcd(int m,int n){return n?gcd(n,m%n):m;}
int lcm(int m,int n){return m/gcd(m,n)*n;}
signed main(){
  int n;
  cin>>n;
  vector<int> x(n);
  for(auto& i:x)cin>>i;
  vector<int> a(n),d(n),f(n);
  f[1]= a[1] = 1;
  for(int i=2;i<n;++i) f[i] = i*f[i-1]%mod;
  for(int i=1;i+1<n;++i) a[i+1] = ((i+1)*a[i]%mod+f[i])%mod;
  for(int i=1;i<n;++i) d[i] = x[i] - x[i-1];
  int res=0;
  for(int i=1;i<=n-2;++i){
    res = (res + a[i]*d[i]%mod)%mod;
    res = (res*(i+1))%mod;
  }
  res = (res + a[n-1]*d[n-1]%mod)%mod;
  cout<<res;
}