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;
typedef long long ll;

int main() {
    // your code goes here
    ll i,n,x[100005],ans=0,mod=1e9+7,d,mm,f=1,seq=1;
    cin >> n;
    for(i=0;i<n;i++){
        cin >> x[i];
        if(i!=0){
            if(i!=1) f=(f*(i-1))%mod;
            d=x[i]-x[i-1];
            ans=(ans*i)%mod;
            if(i!=1){
                seq=(seq*i)%mod;
                seq=(seq+f)%mod;
            }
            mm=(seq*d)%mod;
            ans=(ans+mm)%mod;
        }
    }
    cout << ans << endl;
    return 0;
}