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 <string>
#define REP(i,n) for(int i=0;i<(n);++i)
#define MOD 1000000007
using namespace std;
typedef long long LL;

int main(){
  int N;
  LL _X[100000], X[100000];
  cin >> N;
  REP(i, N) {
    cin >> _X[i];
  }
  REP(i, N) {
    X[i] = _X[N-1] - _X[N-1-i];
  }
  
  LL kaijo[100001], rkaijo[100001];
  kaijo[0] = 1; rkaijo[N] = 1;
  REP(i, N) {
    kaijo[i+1] = kaijo[i] * (i+1) % MOD;
    rkaijo[N-i-1] = rkaijo[N-i] * (N-i-1) % MOD;
  }
  REP(i, N) {
    
  }
  
  LL ret = 0;
  for(int i = 1; i<N; ++i) {
    LL coff = kaijo[N-i-1] * rkaijo[N-i+1] % MOD;
    ret = (ret + X[i] * coff) % MOD;
  }
  cout << ret << endl;
}