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;

long long MOD=1000000007;
long long expo(long long a,long long b){
	if(b==0)return 1;
	long long temp=expo(a,b/2);
	if(b%2==0)return (temp*temp)%MOD;
	return (((temp*temp)%MOD)*a)%MOD;
}

int main(){
	int n;
	scanf("%d",&n);
	long long x[n];
	for(int i=0;i<n;i++)scanf("%lld",&x[i]);
	long long fac[n];
	fac[0]=1;
	for(long long i=1;i<n;i++)fac[i]=(fac[i-1]*i)%MOD;
	long long p[n];
	for(int i=0;i<n;i++){
		if(i==0)p[i]=1;
		else p[i]=p[i-1]+expo(i+1,MOD-2);
		p[i]%=MOD;
	}
	long long ans=0;
	for(int i=0;i<n-1;i++){
		ans+=p[i]*(x[i+1]-x[i]);
		ans%=MOD;
	}
	printf("%lld",(((ans*fac[n-1])%MOD)+MOD)%MOD);
}