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 math
import sys 
input = sys.stdin.readline

N = int(input())
x = list(map(int,input().split()))
mod = 10**9+7

d = [0]*(N-1)

for i in range(N-1):
    d[i] = x[i+1]-x[i]
    d[i] %= mod


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


ans=0
x=0

def pow_r(x, n):
    """
    O(log n)
    """
    x %= mod
    if n == 0:  # exit case
        return 1
    if n % 2 == 0:  # standard case ① n is even
        return pow_r(x ** 2, n // 2)%mod
    else:  # standard case ② n is odd
        return (x * (pow_r(x ** 2, (n - 1) // 2)))%mod

for i in range(N-1):
    #(i+1)%(mod-2)を計算してyへ
    y = pow_r(i+1,mod-2)
    #x+= kaijou//(i+1))%(10**9+7)
    x += kaijou*y
    ans += d[i]*x
    ans %= mod

print(ans)
