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>
#include<cstdio>
#include<cstring>
using namespace std;
#define N 100010
const int mod=1e9+7;
inline int read(){
	int x=0,f=1;
	char c=getchar();
	while(c<'0'||c>'9'){
		if(c=='-')f=-1;
		c=getchar();
	}
	while(c>='0'&&c<='9'){
		x=(x<<1)+(x<<3)+c-'0';
		c=getchar();
	}
	return x*f;
}
int n,a[N],inv[N],ans,sum;
int main(){
	n=read();
	inv[1]=1;
	for(int i=2;i<=n;++i){
		inv[i]=1LL*(mod-mod/i)*inv[mod%i]%mod;
	}
	for(int i=1;i<=n;++i){
		a[i]=read();
	}
	for(int i=1;i<=n;++i){
		ans=(ans+1LL*sum*(a[i]-a[i-1]))%mod;
		sum=(sum+inv[i])%mod;
	}
	for(int i=1;i<n;++i){
		ans=1LL*ans*i%mod;
	}
	printf("%d\n",ans);
	return 0;
}

