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;
#define int long long 

const int MAX_N = 100100;
const int MOD = 1e9+7;
int inv[MAX_N]; 

int p(int a, int n ){ 
  if( n == 0 ){ return 1LL;}
  else if( n == 1 ){ return a;}
  else{
    if( n % 2 == 0 ){ return ( p(a,n/2)*p(a,n/2) )%MOD;}
    else{ return ( ( ( p(a,(n-1)/2 ) * p(a,(n-1)/2) )%MOD ) *a )%MOD;}
  }
}

void init(){
  inv[1] = 1;
  for(int i = 2; i < MAX_N; i++){
    inv[i] = (inv[i-1]+p(i,MOD-2))%MOD;
  }
}

signed main(void){ 
  
  int N; cin >> N; init();
  vector<int> X(N);
  for(int i = 0; i < N; i++){ cin >> X[i];}
  int ans = 0;
  for(int i = 1; i < N; i++){
    int d = (X[i]-X[i-1])%MOD;
    ans += d*inv[i];
    ans %= MOD;
  }
  
  for(int i = 1; i < N; i++){ 
    ans *= i;
    ans %= MOD;
  }
  
  cout << ans << endl;
  
  return 0;
}
   