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<cstring>
#include<algorithm>
#include<cstdio>
#define rep(a,b,c) for (int a=b;a<=c;a++)
#define per(a,b,c) for (int a=b;a>=c;a--)
typedef long long LL;
typedef double dob;
using namespace std;
const int P=1000000007;
const int N=100010;
int n,x[N],d[N],di[N],pi[N],f[N],ans;
int main(){
	scanf("%d",&n);
	rep(i,1,n) scanf("%d",x+i);
	rep(i,1,n-1) d[i]=x[i+1]-x[i];
	pi[0]=1;
	rep(i,1,n) pi[i]=(LL)pi[i-1]*i%P;
	di[n]=1;
	per(i,n-1,1) di[i]=(LL)di[i+1]*i%P;
	rep(i,1,n-1){
		f[i]=((LL)f[i-1]*i%P+pi[i-1])%P;
	}
	rep(i,1,n-1){
		ans=(ans+((LL)d[i]*f[i]%P*di[i+1])%P)%P;
	}
	ans=(ans%P+P)%P;
	printf("%d\n",ans);
	return 0;
}
