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 rep(i,n) for (long long i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<ll,ll>;
using vec = vector<ll>;
using vecp = vector<P>;
using mat = vector<vec>;
using matp = vector<vecp>;
const ll MOD = 1e9+7;
const ll INF = 1e18;
#define all(v) v.begin(), v.end()

ll modPow(ll a, ll n) {
  if (n == 0) return 1; // 0乗にも対応する場合
  if (n == 1) return a % MOD;
  if (n % 2 == 1) return (a * modPow(a, n - 1)) % MOD;
  ll t = modPow(a, n / 2);
  return (t * t) % MOD;
}

int main(){
  ll N,ans=0,x=1;
  cin >> N;
  vec X(N),P(N);
  rep(i,N-1){
    x*=i+1;
    x%=MOD;
  }
  rep(i,N){
    if(i==0){
      P.at(i)=x;
    }else{
      P.at(i)=P.at(i-1)+x*modPow(i+1,MOD-2);
      P.at(i)%=MOD;
    }
  }
  cin >> X.at(0);
  rep(i,N-1){
    cin >> X.at(i+1);
    ans+=(X.at(i+1)-X.at(i))*P.at(i);
    ans%=MOD;
  }
  cout << ans << endl;
}