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;

const int MAX_N = 100100;
const ll MOD = 1e9+7;

ll pMOD( ll a, ll p ){ 
  // aのp乗数をMODで返す
  
  if( p == 0 ){ return 1LL;}
  else if( p == 1 ){ return a;}
  else{
    
    if( p % 2 == 0 ){ return ( pMOD(a,p/2)*pMOD(a,p/2) ) % MOD;}
    else{
      return ( a * ( ( pMOD( a, (p-1)/2 ) * pMOD( a, (p-1)/2 ) )%MOD ) ) % MOD;
    }
    
  }
}

int main(){ 
  
  ll N; cin >> N;
  ll x[MAX_N];
  
  for(ll i = 1; i <= N; i++){ cin >> x[i];}
  
  ll ex = 1LL;
  
  for(ll i = 1; i < N; i++){ ex *= i; ex %= MOD;}
  
  // exは(N-1)! % MOD
  
  ll b[MAX_N];
  
  b[1] = ex;
  
  for(ll i = 1; i+1 < N; i++){ 
    
    b[i+1] = b[i]+ex*pMOD( i+1, MOD-2 );
    
    b[i+1] %= MOD;
    
  }
  
  ll ans = 0;
  
  for(ll i = 1; i < N; i++){ 
    
    ans += (x[i+1]-x[i]) * b[i];
    
    ans %= MOD;
    
  }
  
  
  cout << ans << endl;
  
  return 0;
} 
  
  