#!/usr/bin/env python3

import sys
from typing import Any, Callable, Deque, Dict, List, Mapping, Optional, Sequence, Set, Tuple, TypeVar, Union
# import time
# import math, cmath
# import numpy as np
# import scipy.sparse.csgraph as cs            # csgraph_from_dense(ndarray, null_value=inf), bellman_ford(G, return_predecessors=True), dijkstra, floyd_warshall
# import random                                # random, uniform, randint, randrange, shuffle, sample
# import string                                # ascii_lowercase, ascii_uppercase, ascii_letters, digits, hexdigits
# import re                                    # re.compile(pattern) => ptn obj; p.search(s), p.match(s), p.finditer(s) => match obj; p.sub(after, s)
# from bisect import bisect_left, bisect_right # bisect_left(a, x, lo=0, hi=len(a)) returns i such that all(val<x for val in a[lo:i]) and all(val>-=x for val in a[i:hi]).
# from collections import deque                # deque class. deque(L): dq.append(x), dq.appendleft(x), dq.pop(), dq.popleft(), dq.rotate()
# from collections import defaultdict          # subclass of dict. defaultdict(facroty)
# from collections import Counter              # subclass of dict. Counter(iter): c.elements(), c.most_common(n), c.subtract(iter)
# from datetime import date, datetime          # date.today(), date(year,month,day) => date obj; datetime.now(), datetime(year,month,day,hour,second,microsecond) => datetime obj; subtraction => timedelta obj
# from datetime.datetime import strptime       # strptime('2019/01/01 10:05:20', '%Y/%m/%d/ %H:%M:%S') returns datetime obj
# from datetime import timedelta               # td.days, td.seconds, td.microseconds, td.total_seconds(). abs function is also available.
# from copy import copy, deepcopy              # use deepcopy to copy multi-dimentional matrix without reference
# from functools import reduce                 # reduce(f, iter[, init])
# from functools import lru_cache              # @lrucache ...arguments of functions should be able to be keys of dict (e.g. list is not allowed)
# from heapq import heapify, heappush, heappop # built-in list. heapify(L) changes list in-place to min-heap in O(n), heappush(heapL, x) and heappop(heapL) in O(lgn).
# from heapq import _heapify_max, _heappop_max, _siftdown_max
# from heapq import nlargest, nsmallest        # nlargest(n, iter[, key]) returns k-largest-list in O(n+klgn).
# from itertools import count, cycle, repeat   # count(start[,step]), cycle(iter), repeat(elm[,n])
# from itertools import groupby                # [(k, list(g)) for k, g in groupby('000112')] returns [('0',['0','0','0']), ('1',['1','1']), ('2',['2'])]
# from itertools import starmap                # starmap(pow, [[2,5], [3,2]]) returns [32, 9]
# from itertools import product, permutations  # product(iter, repeat=n), permutations(iter[,r])
# from itertools import combinations, combinations_with_replacement
from itertools import accumulate             # accumulate(iter[, f])
# from operator import itemgetter              # itemgetter(1), itemgetter('key')
# from fractions import Fraction               # Fraction(a, b) => a / b ∈ Q. note: Fraction(0.1) do not returns Fraciton(1, 10). Fraction('0.1') returns Fraction(1, 10)



def main():
    Num = Union[int, float]
    mod = 1000000007                # 10^9+7
    inf = float('inf')              # sys.float_info.max = 1.79e+308
    # inf = 2 ** 63 - 1             # (for fast JIT compile in PyPy) 9.22e+18
    sys.setrecursionlimit(10**6)    # 1000 -> 1000000
    def input(): return sys.stdin.readline().rstrip()
    def ii():    return int(input())
    def isp():   return input().split()
    def mi():    return map(int, input().split())
    def mi_0():  return map(lambda x: int(x)-1, input().split())
    def lmi():   return list(map(int, input().split()))
    def lmi_0(): return list(map(lambda x: int(x)-1, input().split()))
    def li():    return list(input())
    # def _heappush_max(h, item): h.append(item); _siftdown_max(h, 0, len(h)-1)


    class Combinatorics:
        def __init__(self, size: int, mod: int):
            '''
            >>> c = Combinatorics(size=10, mod=10**9+7)
            >>> c.fact
            [1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800]
            >>> c.inv
            [1, 1, 500000004, 166666668, 41666667, 808333339, 301388891, 900198419, 487524805, 831947206, 283194722]
            '''
            self.size =  size
            self.mod = mod
            self.fact = self._build_factorial(size, mod)
            self.inv = self._build_inv_factorial(size, mod)
        
        def _build_factorial(self, size:int , mod: int) -> None:
            '''
            fact_mod[i] は i! % mod を表すとする
            O(n) で fact_mod[size] まで計算結果がメモされたテーブルを構築して返す
            '''
            fact = [1] * (size + 1)
            for i in range(1, size + 1):
                fact[i] = (fact[i-1] * i) % mod
            return fact
        
        def _build_inv_factorial(self, size: int, mod: int) -> None:
            '''
            inv_fact_mod[i] は 1/i! % mod を表すとする
            O(n+lgm) で inv_fact_mod[size] まで計算結果がメモされたテーブルを構築して返す
            Note:
                [フェルマーの小定理]
                a ^ p-1 ≡ 1 (mod p)
                a ^ p-2 ≡ 1/a (mod p) (逆元)
                nCr = (n!) / ((n-r)! * r!) だが、mod p の世界ではこの分母を逆元を用いて計算しておくことが可能
            '''
            inv_fact_mod = [1] * (size + 1)
            inv_fact_mod[size] = pow(self.fact[size], mod-2, mod)    # a ^ p-2 ≡ 1/a (mod p) において a = n! とする。 1/n! (mod p) を求める
            for i in range(size-1, -1, -1):
                inv_fact_mod[i] = (inv_fact_mod[i+1] * (i+1)) % mod    # 1/(n-1)! = 1/n! * n
            return inv_fact_mod
        
        def comb(self, n: int, r: int) -> int:
            '''
            nCr % mod を O(1) で計算する。
            Examples:
                >>> c = Combinatorics(size=100, mod=10**9+7)
                >>> c.comb(100, 50)
                538992043   
            '''
            numerator = self.fact[n]
            denominator = (self.inv[n-r] * self.inv[r]) % self.mod
            return (numerator * denominator) % self.mod
    

    n = ii()
    L = lmi()¥

    c = Combinatorics(10**5, mod)
    seq = [(c.fact[n-1] * pow(i, mod-2, mod)) % mod for i in range(1, n)]
    ac = list(accumulate(seq))

    ans = 0
    for i in range(n-1):
        tmp = (L[i+1] - L[i]) * ac[i] % mod
        ans = (ans + tmp) % mod
    print(ans)


if __name__ == "__main__":
    main()


### FIXED CODE ###

#!/usr/bin/env python3

import sys
from typing import Any, Callable, Deque, Dict, List, Mapping, Optional, Sequence, Set, Tuple, TypeVar, Union
# import time
# import math, cmath
# import numpy as np
# import scipy.sparse.csgraph as cs            # csgraph_from_dense(ndarray, null_value=inf), bellman_ford(G, return_predecessors=True), dijkstra, floyd_warshall
# import random                                # random, uniform, randint, randrange, shuffle, sample
# import string                                # ascii_lowercase, ascii_uppercase, ascii_letters, digits, hexdigits
# import re                                    # re.compile(pattern) => ptn obj; p.search(s), p.match(s), p.finditer(s) => match obj; p.sub(after, s)
# from bisect import bisect_left, bisect_right # bisect_left(a, x, lo=0, hi=len(a)) returns i such that all(val<x for val in a[lo:i]) and all(val>-=x for val in a[i:hi]).
# from collections import deque                # deque class. deque(L): dq.append(x), dq.appendleft(x), dq.pop(), dq.popleft(), dq.rotate()
# from collections import defaultdict          # subclass of dict. defaultdict(facroty)
# from collections import Counter              # subclass of dict. Counter(iter): c.elements(), c.most_common(n), c.subtract(iter)
# from datetime import date, datetime          # date.today(), date(year,month,day) => date obj; datetime.now(), datetime(year,month,day,hour,second,microsecond) => datetime obj; subtraction => timedelta obj
# from datetime.datetime import strptime       # strptime('2019/01/01 10:05:20', '%Y/%m/%d/ %H:%M:%S') returns datetime obj
# from datetime import timedelta               # td.days, td.seconds, td.microseconds, td.total_seconds(). abs function is also available.
# from copy import copy, deepcopy              # use deepcopy to copy multi-dimentional matrix without reference
# from functools import reduce                 # reduce(f, iter[, init])
# from functools import lru_cache              # @lrucache ...arguments of functions should be able to be keys of dict (e.g. list is not allowed)
# from heapq import heapify, heappush, heappop # built-in list. heapify(L) changes list in-place to min-heap in O(n), heappush(heapL, x) and heappop(heapL) in O(lgn).
# from heapq import _heapify_max, _heappop_max, _siftdown_max
# from heapq import nlargest, nsmallest        # nlargest(n, iter[, key]) returns k-largest-list in O(n+klgn).
# from itertools import count, cycle, repeat   # count(start[,step]), cycle(iter), repeat(elm[,n])
# from itertools import groupby                # [(k, list(g)) for k, g in groupby('000112')] returns [('0',['0','0','0']), ('1',['1','1']), ('2',['2'])]
# from itertools import starmap                # starmap(pow, [[2,5], [3,2]]) returns [32, 9]
# from itertools import product, permutations  # product(iter, repeat=n), permutations(iter[,r])
# from itertools import combinations, combinations_with_replacement
from itertools import accumulate             # accumulate(iter[, f])
# from operator import itemgetter              # itemgetter(1), itemgetter('key')
# from fractions import Fraction               # Fraction(a, b) => a / b ∈ Q. note: Fraction(0.1) do not returns Fraciton(1, 10). Fraction('0.1') returns Fraction(1, 10)



def main():
    Num = Union[int, float]
    mod = 1000000007                # 10^9+7
    inf = float('inf')              # sys.float_info.max = 1.79e+308
    # inf = 2 ** 63 - 1             # (for fast JIT compile in PyPy) 9.22e+18
    sys.setrecursionlimit(10**6)    # 1000 -> 1000000
    def input(): return sys.stdin.readline().rstrip()
    def ii():    return int(input())
    def isp():   return input().split()
    def mi():    return map(int, input().split())
    def mi_0():  return map(lambda x: int(x)-1, input().split())
    def lmi():   return list(map(int, input().split()))
    def lmi_0(): return list(map(lambda x: int(x)-1, input().split()))
    def li():    return list(input())
    # def _heappush_max(h, item): h.append(item); _siftdown_max(h, 0, len(h)-1)


    class Combinatorics:
        def __init__(self, size: int, mod: int):
            '''
            >>> c = Combinatorics(size=10, mod=10**9+7)
            >>> c.fact
            [1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800]
            >>> c.inv
            [1, 1, 500000004, 166666668, 41666667, 808333339, 301388891, 900198419, 487524805, 831947206, 283194722]
            '''
            self.size =  size
            self.mod = mod
            self.fact = self._build_factorial(size, mod)
            self.inv = self._build_inv_factorial(size, mod)
        
        def _build_factorial(self, size:int , mod: int) -> None:
            '''
            fact_mod[i] は i! % mod を表すとする
            O(n) で fact_mod[size] まで計算結果がメモされたテーブルを構築して返す
            '''
            fact = [1] * (size + 1)
            for i in range(1, size + 1):
                fact[i] = (fact[i-1] * i) % mod
            return fact
        
        def _build_inv_factorial(self, size: int, mod: int) -> None:
            '''
            inv_fact_mod[i] は 1/i! % mod を表すとする
            O(n+lgm) で inv_fact_mod[size] まで計算結果がメモされたテーブルを構築して返す
            Note:
                [フェルマーの小定理]
                a ^ p-1 ≡ 1 (mod p)
                a ^ p-2 ≡ 1/a (mod p) (逆元)
                nCr = (n!) / ((n-r)! * r!) だが、mod p の世界ではこの分母を逆元を用いて計算しておくことが可能
            '''
            inv_fact_mod = [1] * (size + 1)
            inv_fact_mod[size] = pow(self.fact[size], mod-2, mod)    # a ^ p-2 ≡ 1/a (mod p) において a = n! とする。 1/n! (mod p) を求める
            for i in range(size-1, -1, -1):
                inv_fact_mod[i] = (inv_fact_mod[i+1] * (i+1)) % mod    # 1/(n-1)! = 1/n! * n
            return inv_fact_mod
        
        def comb(self, n: int, r: int) -> int:
            '''
            nCr % mod を O(1) で計算する。
            Examples:
                >>> c = Combinatorics(size=100, mod=10**9+7)
                >>> c.comb(100, 50)
                538992043   
            '''
            numerator = self.fact[n]
            denominator = (self.inv[n-r] * self.inv[r]) % self.mod
            return (numerator * denominator) % self.mod
    

    n = ii()
    L = lmi()

    c = Combinatorics(10**5, mod)
    seq = [(c.fact[n-1] * pow(i, mod-2, mod)) % mod for i in range(1, n)]
    ac = list(accumulate(seq))

    ans = 0
    for i in range(n-1):
        tmp = (L[i+1] - L[i]) * ac[i] % mod
        ans = (ans + tmp) % mod
    print(ans)


if __name__ == "__main__":
    main()