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 N = 1e5 + 5, mod = 1e9 + 7;

int x[N];

long long powmod(long long b, long long p) {
  long long r = 1;
  for (; p; p >>= 1, b = b * b % mod)
    if (p & 1)
      r = r * b % mod;
  return r;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", x+i);
  }
  for (int i = 0; i + 1 < n; ++i) {
    x[i] = x[i+1] - x[i];
  }
  long long fact = 1;
  --n;
  for (int i = 1; i <= n; ++i)
    fact = fact * i % mod;
  long long now = 0, ans = 0;
  for (int i = 0; i < n; ++i) {
    now = (now + fact * powmod(i+1, mod-2)) % mod;
    ans = (ans + now * x[i]) % mod;
  }
  printf("%lld\n", ans);
  return 0;
}
