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 ll long long
using namespace std;
int n;
ll mod=1000000007;
ll fac[100010],revfac[100010],mul[100010],sum[100010],a[100010],ans;
int main()
{
    scanf("%d",&n);
    fac[0]=1;
    for(int i=1;i<=n;i++)
    {
        scanf("%lld",a+i);
        fac[i]=fac[i-1]*i;
        fac[i]%=mod;
    }
    revfac[n]=1;
    for(int i=n-1;i>0;i--)
    {
        revfac[i]=i*revfac[i+1];
        revfac[i]%=mod;
    }
    for(int i=1;i<n;i++)
    {
        mul[i]=fac[i-1]*revfac[i+1];
        mul[i]%=mod;
        sum[i]=sum[i-1]+mul[i];
        sum[i]%=mod;
    }
    for(int i=1;i<n;i++)
    {
        ans+=(a[i+1]-a[i])*sum[i];
        ans%=mod;
    }
    printf("%lld",ans);
}