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;
#define rep(i,n) for(int i = 0; i < (int)n; i++)
using ll = long long;
const ll mod = 1e9+7;

void extgcd(ll a, ll b, ll &x, ll &y) {
  if(b == 0) {
    x = 1;
    y = 0;
    return;
  }
  extgcd(b, a%b, y, x);
  y -= a/b*x;
  return; 
}

ll mod_inverse(ll a, ll m) {
  ll x, y;
  extgcd(a,m,x,y);
  return (m + x % m) % m;
}

int main(){
  int n;
  cin >> n;
  ll x[n];
  rep(i,n) cin >> x[i];
  ll ans = 0;
  ll s = 0;
  for(ll i = 1; i < n; i++) {
    s += mod_inverse(i,mod);
    (ans += (x[i]-x[i-1])*s) %= mod;
  }
  for(ll i = 1; i < n; i++) (ans *= i) %= mod;
  cout << ans << endl;
  return 0;
}