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, n) for (ll i = 0; i < (ll)(n); i++)
ll ans = 0;
const ll mo = 1000000007;
auto mod = [](ll modx) {modx%=mo;modx+=mo;modx%=mo;return modx; };
vector<ll> factorial(ll n)
{
  vector<ll> fac;
  fac.push_back(1);
  for (ll i = 1; i < n + 10; i++)
  {
    fac.push_back((fac[i - 1] * i) % mo);
  }
  return fac;
}
ll mypow(ll a, ll b)
{
  ll res = 1;
  a %= mo;
  while (b)
  {
    if (b & 1)
      res = res * a % mo;
    a = a * a % mo;
    b >>= 1;
  }
  return res;
}
int main()
{
ll n ;
cin >> n ;
vector<ll> a(n);
rep(i, n){
  cin >> a[i];
}
vector<ll> fac = factorial(100000);
ll sum=fac[n-1];
rep(i,n-1){
  ans=mod(ans+(a[i+1]-a[i])*sum);
  sum=mod(sum+fac[n-1]*mypow(i+2,mo-2));
}
cout << ans << endl;
return 0;
}