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 <vector>
#include <algorithm>

using namespace std;

typedef long long ll;

const ll mod = 1e9 + 7;
ll n;
ll fac[100005];
ll rfac[100005];
ll ans;
ll x[100005];
ll f(ll a, ll x) {
  ll res = 1;
  while(x > 0) {
    if(x & 1) res = res * a % mod;
    a = a * a % mod;
    x >>= 1;
  }
  return res;
}

int main() {
  cin >> n;

  for(int i = 0; i < n; i++) cin >> x[i];

  fac[0] = 1;
  
  for(ll i = 1; i <= n; i++) {
    fac[i] = fac[i-1] * i % mod;
  }
  for(ll i = 1; i <= n; i++) rfac[i] = f(i,mod-2);
  ll tmp = 0;
  
  for(ll i = 1; i < n; i++) {
    tmp += fac[n-1] * rfac[i] % mod;
    tmp %= mod;
    ans = (ans + tmp * (x[i] - x[i-1]) % mod) % mod;
  }
   
    // cout << fac[n-1] * rfac[i] % mod << endl;
  

  cout << ans << endl;
			       

  
}
