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;
int main(void){
    long long n,x[100010],y[100010],mod=1000000000+7,k[100010],h1[100010],h2[100010];
    cin>>n;
    for(int i=0;i<n;i++) {
        cin>>x[i];
        k[i]=1;
    }
    for(int i=0;i<n+5;i++){
        h2[i]=1;
        h1[i]=1;
    }
    for(int i=1;i<n;i++){
        y[i]=x[i]-x[i-1];
    }
    for(int i=1;i<=n-1;i++){
        h1[i]=(h1[i-1]*i)%mod;
    }
    for(int i=n-1;i>=1;i--){
        h2[i]=(h2[i+1]*i)%mod;
    }
    for(int i=1;i<=n-1;i++){
        k[i]=(h1[i-1]*h2[i+1])%mod;
    }
    long long ans=0,b=0;
    for(int i=1;i<n;i++){
        b=(b+k[i])%mod;
        ans=(ans+(b*y[i])%mod)%mod;
    }
    cout<<ans;
}