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>
#define MOD 1000000007
using namespace std;
inline int add(int a,int b)
{a+=b;return a>=MOD?a-MOD:a;}
inline int sub(int a,int b)
{a-=b;return a<0?a+MOD:a;}
inline int mul(int a,int b)
{return 1LL*a*b%MOD;}
int f[1<<20],inv[1<<20];
int main()
{
	int n,x,als=0,fac=1;
	scanf("%d",&n);
	inv[1]=1;
	int si=0;
	for(int i=2;i<=n;i++)inv[i]=mul(inv[MOD%i],MOD-MOD/i);
	for(int i=1;i<=n;i++)
		f[i]=add(mul(inv[i],inv[i+1]),f[i-1]),fac=i==n?fac:mul(fac,i);
	for(int i=1;i<n;i++)
		scanf("%d",&x),als=add(als,mul(x,sub(f[i-1],add(f[n-i-1],inv[n-i])))),si=add(si,inv[i]);
	scanf("%d",&x);
	printf("%d\n",mul(add(als,mul(x,si)),fac));
	return 0;
}