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 <iostream>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <string>
#include <cmath>
using namespace std;
typedef long long ll;
typedef pair<ll,ll>PP;
const ll MOD = 1000000007;
ll T[100005]={0};
ll S[100005]={0};
ll modpow(ll a, ll mod) {
    ll res = 1;
    ll n=MOD-2;
    while (n > 0) {
        if (n & 1) res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}
int main(){
  int n;cin >>n;
  for(int i=1;i<=n;i++){
    cin>>T[i];
  }
  for(int i=1;i<n;i++){
   T[i]=T[i+1]-T[i];
  }
  for(int i=1;i<=n;i++){
    S[i]=modpow(i,MOD);
  }
  //cout<<" "<<S[1]<<S[2]<<endl;
  for(int i=2;i<=n;i++){
    S[i]=(S[i-1]+S[i])%MOD;
  }
  ll ans=0;
  for(int i=1;i<n;i++){
    ans+=(T[i]*S[i])%MOD;
    ans%=MOD;
  }
  for(int i=1;i<n;i++){
    ans*=i;
    ans%=MOD;
  }
  cout<<ans<<endl;
}
