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;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rept(i,k,n) for(int i = (k); i < (int)(n); i++)
#define all(v) v.begin(), v.end()
#define Sort(v) sort(v.begin(),v.end())
#define Reverse(v) reverse(v.begin(),v.end())

typedef long long ll;
typedef pair<int,int> Pii;
typedef pair<long long,long long> Pdd;
 

int main(){
    ll n,mod=1000000007;
    cin>>n;
    vector<ll> dist(n-1);
    ll a,b;cin>>a;
    rep(i,n-1){
        cin>>b;
        dist[i]=b-a;
        a=b;
    }
    ll ans=dist[0];
    ll ex=1,ls=1;
    rept(i,2,n){
        ex=ex*(i-1)%mod;
        ls =(ls*i+ex)%mod;
        ans = (ans*i+dist[i-1]*ls)%mod;
    }
    cout<<ans<<endl;
    return 0;
}