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 N=100005,Mod=1e9+7;
int a[N];
inline int qpow(int a,int b){
	int ans=1;
	while (b){
		if (b&1) ans=1ll*ans*a%Mod;
		a=1ll*a*a%Mod,b>>=1;
	}
	return ans;
}
int f[N],fac[N];
int main (){
	int n;scanf ("%d",&n);
	for (int i=1;i<=n;i++) scanf ("%d",&a[i]);fac[0]=1;
	for (int i=1;i<=n;i++) fac[i]=1ll*fac[i-1]*i%Mod;
	int now=0,p=0;
	for (int i=2;i<=n;i++){
		p=(p+qpow(i-1,Mod-2))%Mod;
		now=(now+1ll*p*(a[i]-a[i-1]))%Mod;
	}
	printf ("%d",1ll*now*fac[n-1]%Mod);
	return 0;
}
