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;

const int MOD = 1000000007;

int add(int x, int y) { return ((x += y) >= MOD) ? x - MOD : x; }
int mul(int x, int y) { return (long long) x * y % MOD; }
int pw(int x, int y) {
   int ans = 1;
   for (; y > 0; y >>= 1, x = mul(x, x)) {
      if (y & 1) ans = mul(ans, x);
   }
   return ans;
}

int main() {
   ios_base::sync_with_stdio(false);
   int N;
   cin >> N;
   vector<int> X(N);
   for (int i = 0; i < N; ++i) {
      cin >> X[i];
   }
   int coeff = 0;
   int ans = 0;
   int fact = 1;
   for (int i = 0; i + 1 < N; ++i) {
      coeff = add(coeff, pw(i + 1, MOD - 2));
      fact = mul(fact, i + 1);
      ans = add(ans, mul(coeff, X[i + 1] - X[i]));
   }
   cout << mul(ans, fact) << "\n";
   return 0;
}
