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>

typedef long long LL;

const int N=100005;
const int MOD=1000000007;

int n,x[N],ny[N];

int main()
{
	scanf("%d",&n);
	for (int i=1;i<=n;i++) scanf("%d",&x[i]);
	ny[0]=ny[1]=1;
	for (int i=2;i<=n;i++) ny[i]=(LL)(MOD-MOD/i)*ny[MOD%i]%MOD;
	for (int i=2;i<=n;i++) (ny[i]+=ny[i-1])%=MOD;
	int ans=0;
	for (int i=1;i<n;i++) (ans+=(LL)ny[i]*(x[i+1]-x[i])%MOD)%=MOD;
	for (int i=2;i<n;i++) ans=(LL)ans*i%MOD;
	printf("%d\n",ans);
	return 0;
}