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;
long long n,p,arr[100001],fact[100001],tot,dp[100001];
long long power(long long x,long long y){
    long long res=1;
    x%=p;
    while (y>0){
        if (y%2==1) res=(res*x)%p;
        y/=2;
        x=(x*x)%p;
    }
    return res;
}
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cin>>n;
    for (int i=1;i<=n;i++){
        cin>>arr[i];
    }
    p=1e9+7;
    fact[0]=1;
    for (int i=1;i<=n;i++){
        fact[i]=(fact[i-1]*i)%p;
        dp[i]=dp[i-1]+power(i,p-2);
    }
    for (int i=1;i<n;i++){
        long long cur=dp[i];
        tot+=cur*(arr[i+1]-arr[i]);
        tot%=p;
    }
    tot*=fact[n-1];
    tot%=p;
    cout<<tot;
}