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;
typedef vector<ll> vll;

ll mod=1e9+7;

ll mpow(ll x, ll n) {
  ll ans=1;
  while(n>0) {
    if(n&1) {
      ans=ans*x%mod;
    }
    x=x*x%mod;
    n>>=1;
  }
  return ans;
}

vector<ll> tfact(1e7+1,-1);
ll fact(ll n) {
  if(tfact[n]<0) {
    if(n==0) {
      tfact[n]=1;
    }
    else {
      tfact[n]=fact(n-1)*n%mod;
    }
  }
  return tfact[n];
}

int main() {
  ll n;
  cin >> n;
  vll x(n);
  for(ll i;i<n;i++) {
    cin >> x[i];
  }
  vll back(n);
  for(int i=0;i<n;i++) {
    back[i]=mpow(i+1,mod-2);
  }
  vll sback(n);
  for(int i=0;i<n;i++) {
    if(i==0) {
      sback[i]=back[i];
    }
    else {
      sback[i]=(sback[i-1]+back[i])%mod;
    }
  }
  ll ans=0;
  for(int i=0;i<n-1;i++) {
    ans=(ans+(x[i+1]-x[i])*sback[i]%mod)%mod;
  }
  ans=ans*fact(n-1)%mod;
  cout << ans << endl;
}