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

# コンビネーション、さらに高速。あらかじめO(N)の計算をすることでのちの計算が早くなる
def cmb(n,r,mod):
  if (r<0 or r>n):
    return 0
  r=min(r,n-r)
  return g1[n]*g2[r]*g2[n-r]%mod
g1=[1,1] # g1[i]=i! % mod　:階乗
g2=[1,1] # g2[i]=(i!)^(-1) % mod　:階乗の逆元
inverse=[0,1]
for i in range(2,n+1):
  g1.append((g1[-1]*i)%mod)
  inverse.append((-inverse[mod%i]*(mod//i))%mod)
  g2.append((g2[-1]*inverse[-1])%mod)

ans=0
nn=[g1[n-1]]
tmp=g1[n-1]
for i in range(2,n):
  tmp+=g1[n-1]*pow(i,mod-2,mod)
  tmp%=mod
  nn.append(tmp)
  
for i in range(n-1):
  d=x[i+1]-x[i]
  ans+=d*nn[i]
  ans%=mod
print(ans)
