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
import sys

read = sys.stdin.read

N, *x = map(int, read().split())
mod = 10 ** 9 + 7
'''
x_(i+1)が選ばれる前にx_iが選ばれれば, x_(i+1)-x_iを加算
(N-1)!通りの中でこのような状況はsum (N-i)! i=2toN-1

x_(i+n)が選ばれる前、かつx_(i+1)からx_(i+n-1)が選ばれた後にx_iが選ばれれば, x_(i+n)-x_iを加算

'''
inv = [0] * N
for i in range(1, N):
    inv[i] = pow(i, mod - 2, mod)

for i in range(2, N):
    inv[i] = (inv[i] + inv[i - 1]) % mod

fact = 1
for i in range(1, N):
    fact *= i
    fact %= mod

answer = 0
for i, j, k in zip(inv[1:], x, x[1:]):
    tmp = (i * fact) % mod
    answer += tmp * (k - j) % mod
    answer %= mod

print(answer)
