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;
const int mod=1000000007;
inline void Add(int &a,int b){a=a+b>=mod?a+b-mod:a+b;}
inline int ksm(int a,int b)
{
	int res=1;
	for(;b;b>>=1,a=1ll*a*a%mod) if(b&1) res=1ll*res*a%mod;
	return res;
}
int sum=0,n,a[100100],ans=0;
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++) scanf("%d",a+i);
	for(int i=2;i<=n;i++)
	{
		Add(sum,ksm(i-1,mod-2));
		Add(ans,1ll*sum*(a[i]-a[i-1])%mod);
	}
	for(int i=1;i<=n-1;i++) ans=1ll*ans*i%mod;
	cout<<ans<<endl;
}