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
# -*- coding: utf-8 -*-
# dwacon6th-prelims/dwacon6th_prelims_b
import sys

s2nn = lambda s: [int(c) for c in s.split(' ')]
i2s = lambda: sys.stdin.readline().rstrip()
i2n = lambda: int(i2s())
i2nn = lambda: s2nn(i2s())

MAX = 10**5 + 100
MOD = 10**9 + 7

fac = [1, 1] + [0] * MAX
finv = [1, 1] + [0] * MAX
inv = [1, 1] + [0] * MAX
coef = [1, 1] + [0] * MAX

for i in range(2, MAX):
    fac[i] = fac[i - 1] * i % MOD
    inv[i] = MOD - inv[MOD % i] * (MOD // i) % MOD
    finv[i] = finv[i - 1] * inv[i] % MOD
    coef[i] = (coef[i - 1] * i + fac[i - 1]) % MOD

def main():
    N = i2n()
    n = N - 1
    x = i2nn()
    total = 0
    for i in range(1, N):
        d = x[i] - x[i - 1]
        total = (total + coef[i] * fac[n] % MOD * finv[i] % MOD * d % MOD) % MOD
    print(total)
    return

main()
