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

n=int(input())
a=[int(k) for k in input().strip().split(" ")]
ai=a[::-1]
res=0

mod=10**9+7
fact=[1]
for i in range(1,100000):
    fact+=[(i*fact[-1]%mod)]

sa=[0]
for k in a:
    sa+=[(sa[-1]+k)%mod]

def ssum(l,r):
    return (sa[r+1]-sa[l])%mod

b=[[0 for i in range(2)] for j in range(n-1)]
b[0][0]=fact[n-1]
b[0][1]=(b[0][0]*pow(2,mod-2,mod))%mod
res+=(b[0][0]*(a[-1]-a[-2]))%mod
if n>2:
    res+=(b[0][1]*(ssum(1,n-2)-ssum(0,n-3)))%mod
for j in range(2,n):
    b[j-1][0]=(b[j-2][0]*(j-1) * pow(j,mod-2,mod))%mod
    res+=(b[j-1][0]*(a[-1]-a[-j-1]))%mod
    if j<n-1:
        b[j-1][1]=(b[j-2][1]*(j-1) * pow(j+1,mod-2,mod))%mod
        res+=(b[j-1][1]*(ssum(j,n-2)-ssum(0,n-j-2)))%mod

print(res%mod)