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 mp make_pair
#define pb push_back
#define ii pair<int,int>
#define all(x) (x).begin(),(x).end()
#define INF 100000000000000000
#define modulo 1000000007
#define mod 998244353
#define int long long int
using namespace std;
int exp(int a,int x){
      if(x==0)return 1;
      if(x%2)return a*exp(a,x-1)%modulo;
      int t=exp(a,x/2);
      return t*t%modulo;
}
int32_t main(){
      ios_base::sync_with_stdio(false);
      cin.tie(NULL);
   //   freopen("q.gir","r",stdin);
   //   freopen("q.cik","w",stdout);
      int n;
      cin>>n;
      vector<int>arr(n);
      for(int i=0;i<n;i++)cin>>arr[i];
      int p=0,ans=0;
      for(int i=1;i<n;i++){
            p=(p+exp(i,modulo-2))%modulo;
            ans=(ans+p*(arr[i]-arr[i-1]))%modulo;
      }
      for(int i=1;i<=n-1;i++)ans=ans*i%modulo;
      cout<<ans;
}
