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 rep(i,a) for(int i=0;i<(a);i++)
const ll MOD=1000000007;
//const ll MOD=998244353;

long long modpow(long long a, long long n){
  long long r=1;
  while(n){
    if(n&1) r=r*a%MOD;
    a=a*a%MOD;
    n>>=1;
  }
  return r;
}

ll sum[101010];
ll X[101010];
int N;

int main(){
  cin>>N;
  rep(i,N) cin>>X[i];
  ll fac=1;
  for(int i=2;i<N;i++) (fac*=i)%=MOD;
  for(int i=1;i<N;i++){
    sum[i]=(sum[i-1]+fac*modpow(i,MOD-2))%MOD;
  }
  ll ans=0;
  for(int i=0;i<N-1;i++){
    ans=(ans+sum[i+1]*(X[i+1]-X[i]))%MOD;
  }
  cout<<ans<<endl;
  return 0;
}