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;

const ll MOD = 1e9 + 7;

ll powMod(ll x, ll k) {
  if (k == 0) return 1;
  if (k % 2 == 0)
    return powMod(x * x % MOD, k / 2);
  else
    return x * powMod(x, k - 1) % MOD;
}

int main() {
  int N;
  cin >> N;
  vector<ll> x(N), a(N + 1);
  for (int i = 0; i < N; i++) cin >> x[i];

  ll ans = 0, fac = 1;
  for (int i = 1; i <= N - 1; i++) {
    fac *= i;
    fac %= MOD;
  }

  for (int i = 1; i <= N; i++) {
    a[i] = a[i - 1] + fac * powMod(i, MOD - 2);
    a[i] %= MOD;
  }

  for (int i = 0; i < N - 1; i++) {
    ans = (ans + (x[i + 1] - x[i]) * a[i + 1] % MOD) % MOD;
  }
  cout << ans << endl;
}