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 < (n); ++i)
#define ina(x,n) rep(x_i,n) cin>>x[x_i]

typedef long long ll;
typedef vector<ll> vl;

const int MOD = 1e9+7;

ll power(ll a, ll b) {
  ll res = 1, base = a;
  while (b > 0) {
    if (b & 1) (res *= base) %= MOD;
    (base *= base) %= MOD;
    b >>= 1;
  }
  return res;
}

int main() {
  int n; cin >> n;
  vl x(n); ina(x,n);
  ll ans = 0, c = 0;
  rep(i,n-1) {
    c += power(i+1, MOD-2);
    c %= MOD;
    ans += c * (x[i+1] - x[i]);
    ans %= MOD;
  }
  rep(i,n-1) {
    ans *= (i+1);
    ans %= MOD;
  }
  cout << ans << endl;
  return 0;
}
