p02807 Dwango Programming Contest 6th - Fusing Slimes
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
```python
n = int(input())
lst1 = list(map(int,input().split()))
mod = 10**9+7
def pow(n, p):
    res = 1
    while p > 0:
        if p % 2 == 0:
            n = n ** 2 % mod
            p //= 2
        else:
            res = res * n % mod
            p -= 1
    return res % mod
fac = [1]

for i in range(n-1):
    fac.append(fac[-1]*(i+1)%mod)

f = 1
ans = 0
for i in range(n-1):
    ans += ((lst1[-1]-lst1[i])*fac[-1]*pow(i+1,mod-2))%mod
print(ans%mod)
```