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

MAX = 2*10**5+100
MOD = 10**9+7
fact = [0]*MAX #fact[i]: i!
inv = [0]*MAX #inv[i]: iの逆元
finv = [0]*MAX #finv[i]: i!の逆元
fact[0] = 1
fact[1] = 1
finv[0] = 1
finv[1] = 1
inv[1] = 1
    
for i in range(2, MAX):
    fact[i] = fact[i-1]*i%MOD
    inv[i] = MOD-inv[MOD%i]*(MOD//i)%MOD
    finv[i] = finv[i-1]*inv[i]%MOD

def C(n, r):
    if n<r:
        return 0
    if n<0 or r<0:
        return 0
    return fact[n]*(finv[r]*finv[n-r]%MOD)%MOD
    
N = int(input())
x = list(map(int, input().split()))
ans = 0
acc = [0]

for i in range(N-1):
    acc.append((acc[-1]+C(N-1, i+1)*fact[i]*fact[N-1-(i+1)])%MOD)

for i in range(N-1):
    d = x[i+1]-x[i]
    ans += d*acc[i+1]
    ans %= MOD
    """
    for j in range(i+1):
        ans += d*C(N-1, j+1)*fact[j]*fact[N-1-(j+1)]
        ans %= MOD
    """
print(ans)