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<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#define ll long long
#define mod 1000000007
#define maxn 200010
inline ll read()
{
	ll x=0; char c=getchar(),f=1;
	for(;c<'0'||'9'<c;c=getchar())if(c=='-')f=-1;
	for(;'0'<=c&&c<='9';c=getchar())x=x*10+c-'0';
	return x*f;
}
inline void write(ll x)
{
	static int buf[20],len; len=0;
	if(x<0)x=-x,putchar('-');
	for(;x;x/=10)buf[len++]=x%10;
	if(!len)putchar('0');
	else while(len)putchar(buf[--len]+'0');
}
inline void writeln(ll x){write(x); putchar('\n');}
inline void writesp(ll x){write(x); putchar(' ');}
ll inv[maxn],sum[maxn];
int a[maxn];
int n;
int main()
{
	n=read();
	for(int i=1;i<=n;i++)
		a[i]=read();
	inv[1]=1; sum[1]=1;
	for(int i=2;i<=n;i++){
		inv[i]=(mod-mod/i)*inv[mod%i]%mod;
		sum[i]=(sum[i-1]+inv[i])%mod;
	}
	ll ans=0;
	for(int i=1;i<n;i++)
		ans=(ans+(a[i+1]-a[i])*sum[i])%mod;
	for(int i=1;i<n;i++)
		ans=ans*i%mod;
	writeln(ans);
	return 0;
}