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 int long long
using namespace std;
constexpr int mod=1e9+7;
int fact_arr[100000];
int fact(int n){
    if(fact_arr[n]!=0)
        return fact_arr[n];
    if(n==0)
        return fact_arr[n]=1;
    return fact_arr[n]=fact(n-1)*n%mod;
}
int pow(int x,int n){
    int ret=1;
    while(n){
        if(n&1)
            (ret*=x)%=mod;
        (x*=x)%=mod;
        n>>=1;
    }
    return ret;
}
int inv(int n){
    return pow(n,mod-2);
}
signed main(){
    int n,x[100000];
    cin>>n;
    for(int i=0;i<n;i++)
        cin>>x[i];
    int arr[99999];
    for(int i=0;i<n-1;i++)
        arr[i]=fact(n-1)*inv(i+1)%mod;
    for(int i=1;i<n-1;i++)
        (arr[i]+=arr[i-1])%=mod;
    int ans=0;
    for(int i=0;i<n-1;i++)
        (ans+=(x[i+1]-x[i])*arr[i])%=mod;
    cout<<ans<<endl;
}
