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
input = sys.stdin.readline

def I(): return int(input())
def MI(): return map(int, input().split())
def LI(): return list(map(int, input().split()))

def main():
    mod=10**9+7
    N=I()
    x=LI()
    
    #区間を固定して考える．i番目の区間に対して，i番目のスライムが通過する確率は1
    #i-1番目のスライムの確率は1/2，i-2番目のスライムは1/3...
    
    diff=[0]*(N-1)
    for i in range(N-1):
        diff[i]=x[i+1]-x[i]
    
    p1=[1]*N
    for i in range(N-1):
        p1[i+1]=(p1[i]*(i+1))%mod#1から順にNまでかける
        
    a=[0]*N
    for i in range(N):
        #　a[i]は　mod下での　1/(i+1)
        a[i]=pow(i+1,mod-2,mod)
        
        
    S=[0]*(N+1)
    for i in range(N):
        S[i+1]=(S[i]+a[i])%mod
        
    ans=0
    for i in range(N-1):
        ans=(ans+S[i+1]*diff[i]*p1[-1])%mod
        
    print(ans)
    
    
    
    

main()
