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<iostream>
using namespace std;
int N;
long X[1<<17];
long F[1<<17];
long mod=1e9+7;
long power(long a,long b){return b?power(a*a%mod,b/2)*(b%2?a:1)%mod:1;}
int main()
{
	cin>>N;
	F[0]=1;
	for(long i=1;i<=N;i++)F[i]=F[i-1]*i%mod;
	for(int i=1;i<=N;i++)cin>>X[i];
	long ans=0;
	long t=0;
	for(int i=1;i<N;i++)
	{
		(t+=power(i,mod-2))%=mod;
		long now=F[N-1]*t%mod;
		(ans+=now*(X[i+1]-X[i])%mod)%=mod;
	}
	cout<<ans<<endl;
}