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 <algorithm>
#include <vector>
#include <queue>
#include <iomanip>
#include <cmath>
#define REP(i,n) for (int i=0;i<(n);++i)
using namespace std;
using ll = long long;
constexpr ll MOD = 1e9 + 7;


constexpr int MAX = 100010;
ll fac[MAX], inv[MAX], finv[MAX];
ll n, m, a[100010];

int main() {
  ios::sync_with_stdio(false); cin.tie(0);
  cin >> n;
  REP(i, n) cin >> a[i];

  fac[0] = inv[0] = finv[0] = 1;
  fac[1] = inv[1] = finv[1] = 1;
  for(int i=2; i<MAX; ++i){
    fac[i] = fac[i-1] * i % MOD;
    inv[i] = MOD - inv[MOD%i] * (MOD/i) % MOD;
    finv[i] = finv[i-1] * inv[i] % MOD;
  }

  // 左から期待値を決めていくと漸化式で定められる。(みんな右に向かうので)
  // Ei = 1 / n * (Ei-1 + 1) + (n - 1) / n * Ei-1
  ll e = 0, ans = 0;
  for(int i = 0; i < n - 1; ++i) {
    e = (e + inv[i + 1]) % MOD;
    ans = (ans + e * (a[i + 1] - a[i]) % MOD) % MOD;
  }

  cout << ans * fac[n - 1] % MOD;
  return 0;
}

