Buggy python submission (verdict: Time Limit Exceeded):
#!/usr/bin/env python3

import sys
import math
# import re                                    # re.compile(pattern) => ptn obj; p.search(s), p.match(s), p.finditer(s) => match obj; p.sub(after, s)
# 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 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 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                # product(iter, repeat=n)
# from itertools import accumulate             # accumulate(iter[, f])
# 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
# 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 copy import deepcopy                    # to copy multi-dimentional matrix without reference
# from fractions import gcd                    # for Python 3.4


def main():
    mod = 1000000007                  # 10^9+7
    inf = float('inf')
    sys.setrecursionlimit(10**6)    # 1000 -> 1000000
    def input(): return sys.stdin.readline().rstrip()
    def ii():  return int(input())
    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())
    

    class Eratos:
        def __init__(self, num):
            self.table_max = num
            # self.table[i] は i が素数かどうかを示す (bool)
            self.table = [False if i == 0 or i == 1 else True for i in range(num+1)]
            for i in range(2, int(math.sqrt(num)) + 1):
                if self.table[i]:
                    for j in range(i ** 2, num + 1, i):    # i**2 からスタートすることで定数倍高速化できる
                        self.table[j] = False
            # self.table_max 以下の素数を列挙したリスト
            self.prime_numbers = [2] if self.table_max >= 2 else []
            for i in range(3, self.table_max + 1, 2):
                if self.table[i]:
                    self.prime_numbers.append(i)
        
        def is_prime(self, num):
            if num > self.table_max:
                raise ValueError('Eratos.is_prime(): exceed table_max({}). got {}'.format(self.table_max, num))
            return self.table[num]
        
        def prime_factorize(self, num):
            if num > self.table_max:
                raise ValueError('Eratos.prime_factorize(): exceed table_max({}). got {}'.format(self.table_max, num))
            factorized_dict = defaultdict(int)    # 素因数分解の結果を記録する辞書
            candidate_prime_numbers = [i for i in range(2, int(math.sqrt(num)) + 1) if self.is_prime(i)]
            # n について、√n 以下の素数で割り続けると最後には 1 or 素数となる
            # 背理法を考えれば自明 (残された数が √n より上の素数の積であると仮定。これは自明に n を超えるため矛盾)
            for p in candidate_prime_numbers:
                # これ以上調査は無意味
                if num == 1:
                    break
                while num % p == 0:
                    num //= p
                    factorized_dict[p] += 1
            if num != 1:
                factorized_dict[num] = 1
            return factorized_dict


    n = ii()
    L = lmi()
    a_max = max(L)
    # O(AlglgA) で前処理
    eratos = Eratos(a_max)

    # O(n) * O(√A) で LCM を素因数分解した形で保存する
    lcm_dict = eratos.prime_factorize(L[0])
    for i in range(1, n):
        d = eratos.prime_factorize(L[i])
        for k, v in d.items():
            lcm_dict[k] = max(lcm_dict[k], v)
    
    # O(√A)
    lcm_mod = 1
    for k, v in lcm_dict.items():
        lcm_mod = (lcm_mod * pow(k, v, mod)) % mod
    
    # O(n)
    ans = 0
    for a in L:
        # a^p-2 ≡ 1/a
        inverse_element = pow(a, mod-2, mod)
        ans = (ans + lcm_mod * inverse_element) % mod
    
    print(ans)

    

if __name__ == "__main__":
    main()

Fixed submission:
#!/usr/bin/env python3

import sys
import math
from collections import defaultdict


def main():
    mod = 1000000007                  # 10^9+7
    sys.setrecursionlimit(10**6)    # 1000 -> 1000000
    def input(): return sys.stdin.readline().rstrip()
    def ii():  return int(input())
    def lmi(): return list(map(int, input().split()))

    class Eratos:
        def __init__(self, num):
            self.table_max = num
            # self.table[i] は i が素数かどうかを示す (bool)
            self.table = [False if i == 0 or i == 1 else True for i in range(num+1)]
            for i in range(2, int(math.sqrt(num)) + 1):
                if self.table[i]:
                    for j in range(i ** 2, num + 1, i):    # i**2 からスタートすることで定数倍高速化できる
                        self.table[j] = False
            # self.table_max 以下の素数を列挙したリスト
            self.prime_numbers = [2] if self.table_max >= 2 else []
            for i in range(3, self.table_max + 1, 2):
                if self.table[i]:
                    self.prime_numbers.append(i)
        
        def is_prime(self, num):
            if num > self.table_max:
                raise ValueError('Eratos.is_prime(): exceed table_max({}). got {}'.format(self.table_max, num))
            return self.table[num]
        
        def prime_factorize(self, num):
            if num > self.table_max:
                raise ValueError('Eratos.prime_factorize(): exceed table_max({}). got {}'.format(self.table_max, num))
            factorized_dict = defaultdict(int)    # 素因数分解の結果を記録する辞書
            candidate_prime_numbers = [i for i in range(2, int(math.sqrt(num)) + 1) if self.is_prime(i)]
            # n について、√n 以下の素数で割り続けると最後には 1 or 素数となる
            # 背理法を考えれば自明 (残された数が √n より上の素数の積であると仮定。これは自明に n を超えるため矛盾)
            for p in candidate_prime_numbers:
                # これ以上調査は無意味
                if num == 1:
                    break
                while num % p == 0:
                    num //= p
                    factorized_dict[p] += 1
            if num != 1:
                factorized_dict[num] = 1
            return factorized_dict

    # import time
    # s = time.time()

    n = ii()
    L = lmi()
    a_max = max(L)
    # O(AlglgA) で前処理
    eratos = Eratos(a_max)
    # print('preprocessed (Eratostenes)')
    # print(time.time() - s)

    # O(n) * O(√A) で LCM を素因数分解した形で保存する
    lcm_dict = eratos.prime_factorize(L[0])
    for i in range(1, n):
        d = eratos.prime_factorize(L[i])
        for k, v in d.items():
            lcm_dict[k] = max(lcm_dict[k], v)
    # print('prime factorized')
    # print(time.time() - s)
    
    # O(√A)
    lcm_mod = 1
    for k, v in lcm_dict.items():
        lcm_mod = (lcm_mod * pow(k, v, mod)) % mod
    
    # O(n)
    ans = 0
    for a in L:
        # a^p-2 ≡ 1/a
        inverse_element = pow(a, mod-2, mod)
        ans = (ans + lcm_mod * inverse_element) % mod
    
    print(ans)
    # print(time.time() - s)

    

if __name__ == "__main__":
    main()

Buggy python submission (verdict: Time Limit Exceeded):
#!/usr/bin/env python3

import sys
import math
from collections import defaultdict          # subclass of dict. defaultdict(facroty)



def main():
    mod = 1000000007                  # 10^9+7
    inf = float('inf')
    sys.setrecursionlimit(10**6)    # 1000 -> 1000000
    def input(): return sys.stdin.readline().rstrip()
    def ii():  return int(input())
    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())
    


    class Eratos:
        def __init__(self, num):
            self.table_max = num
            # self.table[i] は i が素数かどうかを示す (bool)
            self.table = [False if i == 0 or i == 1 else True for i in range(num+1)]
            for i in range(2, int(math.sqrt(num)) + 1):
                if self.table[i]:
                    for j in range(i ** 2, num + 1, i):    # i**2 からスタートすることで定数倍高速化できる
                        self.table[j] = False
            # self.table_max 以下の素数を列挙したリスト
            self.prime_numbers = [2] if self.table_max >= 2 else []
            for i in range(3, self.table_max + 1, 2):
                if self.table[i]:
                    self.prime_numbers.append(i)
        
        def is_prime(self, num):
            if num > self.table_max:
                raise ValueError('Eratos.is_prime(): exceed table_max({}). got {}'.format(self.table_max, num))
            return self.table[num]
        
        def prime_factorize(self, num):
            if num > self.table_max:
                raise ValueError('Eratos.prime_factorize(): exceed table_max({}). got {}'.format(self.table_max, num))
            factorized_dict = defaultdict(int)    # 素因数分解の結果を記録する辞書
            candidate_prime_numbers = list(filter(lambda x: x <= math.sqrt(num), self.prime_numbers))
            candidate_prime_numbers.reverse()
            while num > 1 and candidate_prime_numbers:
                p = candidate_prime_numbers[-1]
                if num % p == 0:
                    num //= p
                    factorized_dict[p] += 1
                else:
                    candidate_prime_numbers.pop()
            if num != 1:
                # num は素数
                factorized_dict[num] = 1
            return factorized_dict

        
    n = ii()
    L = lmi()
    a_max = max(L)
    # O(nlglgn) で前処理
    eratos = Eratos(a_max)
    # O(√n) で LCM を素因数分解した形で保存する
    lcm_dict = eratos.prime_factorize(L[0])
    for i in range(1, n):
        d = eratos.prime_factorize(L[i])
        for k, v in d.items():
            lcm_dict[k] = max(lcm_dict[k], v)
    
    lcm_mod = 1
    for k, v in lcm_dict.items():
        lcm_mod = (lcm_mod * pow(k, v, mod)) % mod

    ans = 0
    for a in L:
        # a^p-2 ≡ 1/a
        inverse_element = pow(a, mod-2, mod)
        ans = (ans + lcm_mod * inverse_element) % mod
    
    print(ans)

    

if __name__ == "__main__":
    main()

Fixed submission:
#!/usr/bin/env python3

import sys
import math
from collections import defaultdict


def main():
    mod = 1000000007                  # 10^9+7
    sys.setrecursionlimit(10**6)    # 1000 -> 1000000
    def input(): return sys.stdin.readline().rstrip()
    def ii():  return int(input())
    def lmi(): return list(map(int, input().split()))

    class Eratos:
        def __init__(self, num):
            self.table_max = num
            # self.table[i] は i が素数かどうかを示す (bool)
            self.table = [False if i == 0 or i == 1 else True for i in range(num+1)]
            for i in range(2, int(math.sqrt(num)) + 1):
                if self.table[i]:
                    for j in range(i ** 2, num + 1, i):    # i**2 からスタートすることで定数倍高速化できる
                        self.table[j] = False
            # self.table_max 以下の素数を列挙したリスト
            self.prime_numbers = [2] if self.table_max >= 2 else []
            for i in range(3, self.table_max + 1, 2):
                if self.table[i]:
                    self.prime_numbers.append(i)
        
        def is_prime(self, num):
            if num > self.table_max:
                raise ValueError('Eratos.is_prime(): exceed table_max({}). got {}'.format(self.table_max, num))
            return self.table[num]
        
        def prime_factorize(self, num):
            if num > self.table_max:
                raise ValueError('Eratos.prime_factorize(): exceed table_max({}). got {}'.format(self.table_max, num))
            factorized_dict = defaultdict(int)    # 素因数分解の結果を記録する辞書
            candidate_prime_numbers = [i for i in range(2, int(math.sqrt(num)) + 1) if self.is_prime(i)]
            # n について、√n 以下の素数で割り続けると最後には 1 or 素数となる
            # 背理法を考えれば自明 (残された数が √n より上の素数の積であると仮定。これは自明に n を超えるため矛盾)
            for p in candidate_prime_numbers:
                # これ以上調査は無意味
                if num == 1:
                    break
                while num % p == 0:
                    num //= p
                    factorized_dict[p] += 1
            if num != 1:
                factorized_dict[num] = 1
            return factorized_dict

    # import time
    # s = time.time()

    n = ii()
    L = lmi()
    a_max = max(L)
    # O(AlglgA) で前処理
    eratos = Eratos(a_max)
    # print('preprocessed (Eratostenes)')
    # print(time.time() - s)

    # O(n) * O(√A) で LCM を素因数分解した形で保存する
    lcm_dict = eratos.prime_factorize(L[0])
    for i in range(1, n):
        d = eratos.prime_factorize(L[i])
        for k, v in d.items():
            lcm_dict[k] = max(lcm_dict[k], v)
    # print('prime factorized')
    # print(time.time() - s)
    
    # O(√A)
    lcm_mod = 1
    for k, v in lcm_dict.items():
        lcm_mod = (lcm_mod * pow(k, v, mod)) % mod
    
    # O(n)
    ans = 0
    for a in L:
        # a^p-2 ≡ 1/a
        inverse_element = pow(a, mod-2, mod)
        ans = (ans + lcm_mod * inverse_element) % mod
    
    print(ans)
    # print(time.time() - s)

    

if __name__ == "__main__":
    main()

Buggy python submission (verdict: Time Limit Exceeded):
#!/usr/bin/env python3

import sys
import math
# import re                                    # re.compile(pattern) => ptn obj; p.search(s), p.match(s), p.finditer(s) => match obj; p.sub(after, s)
# 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 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 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                # product(iter, repeat=n)
# from itertools import accumulate             # accumulate(iter[, f])
# 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
# 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 copy import deepcopy                    # to copy multi-dimentional matrix without reference
# from fractions import gcd                    # for Python 3.4


def main():
    mod = 1000000007                  # 10^9+7
    inf = float('inf')
    sys.setrecursionlimit(10**6)    # 1000 -> 1000000
    def input(): return sys.stdin.readline().rstrip()
    def ii():  return int(input())
    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())
    


    class Eratos:
        def __init__(self, num):
            self.table_max = num
            # self.table[i] は i が素数かどうかを示す (bool)
            self.table = [False if i == 0 or i == 1 else True for i in range(num+1)]
            for i in range(2, int(math.sqrt(num)) + 1):
                if self.table[i]:
                    for j in range(i ** 2, num + 1, i):    # i**2 からスタートすることで定数倍高速化できる
                        self.table[j] = False
            # self.table_max 以下の素数を列挙したリスト
            self.prime_numbers = [2] if self.table_max >= 2 else []
            for i in range(3, self.table_max + 1, 2):
                if self.table[i]:
                    self.prime_numbers.append(i)
        
        def is_prime(self, num):
            if num > self.table_max:
                raise ValueError('Eratos.is_prime(): exceed table_max({}). got {}'.format(self.table_max, num))
            return self.table[num]
        
        def prime_factorize(self, num):
            if num > self.table_max:
                raise ValueError('Eratos.prime_factorize(): exceed table_max({}). got {}'.format(self.table_max, num))
            factorized_dict = defaultdict(int)    # 素因数分解の結果を記録する辞書
            candidate_prime_numbers = list(filter(lambda x: x <= num, self.prime_numbers))
            candidate_prime_numbers.reverse()
            while num > 1 and candidate_prime_numbers:
                p = candidate_prime_numbers[-1]
                if num % p == 0:
                    num //= p
                    factorized_dict[p] += 1
                else:
                    candidate_prime_numbers.pop()
            return factorized_dict

        
    n = ii()
    L = lmi()
    a_max = max(L)
    # O(nlglgn) で前処理
    eratos = Eratos(a_max)
    # O(√n) で LCM を素因数分解した形で保存する
    lcm_dict = eratos.prime_factorize(L[0])
    for i in range(1, n):
        d = eratos.prime_factorize(L[i])
        for k, v in d.items():
            lcm_dict[k] = max(lcm_dict[k], v)
    
    lcm_mod = 1
    for k, v in lcm_dict.items():
        lcm_mod = (lcm_mod * pow(k, v, mod)) % mod

    ans = 0
    for a in L:
        # a^p-2 ≡ 1/a
        inverse_element = pow(a, mod-2, mod)
        ans = (ans + lcm_mod * inverse_element) % mod
    
    print(ans)

    

if __name__ == "__main__":
    main()

Fixed submission:
#!/usr/bin/env python3

import sys
import math
from collections import defaultdict


def main():
    mod = 1000000007                  # 10^9+7
    sys.setrecursionlimit(10**6)    # 1000 -> 1000000
    def input(): return sys.stdin.readline().rstrip()
    def ii():  return int(input())
    def lmi(): return list(map(int, input().split()))

    class Eratos:
        def __init__(self, num):
            self.table_max = num
            # self.table[i] は i が素数かどうかを示す (bool)
            self.table = [False if i == 0 or i == 1 else True for i in range(num+1)]
            for i in range(2, int(math.sqrt(num)) + 1):
                if self.table[i]:
                    for j in range(i ** 2, num + 1, i):    # i**2 からスタートすることで定数倍高速化できる
                        self.table[j] = False
            # self.table_max 以下の素数を列挙したリスト
            self.prime_numbers = [2] if self.table_max >= 2 else []
            for i in range(3, self.table_max + 1, 2):
                if self.table[i]:
                    self.prime_numbers.append(i)
        
        def is_prime(self, num):
            if num > self.table_max:
                raise ValueError('Eratos.is_prime(): exceed table_max({}). got {}'.format(self.table_max, num))
            return self.table[num]
        
        def prime_factorize(self, num):
            if num > self.table_max:
                raise ValueError('Eratos.prime_factorize(): exceed table_max({}). got {}'.format(self.table_max, num))
            factorized_dict = defaultdict(int)    # 素因数分解の結果を記録する辞書
            candidate_prime_numbers = [i for i in range(2, int(math.sqrt(num)) + 1) if self.is_prime(i)]
            # n について、√n 以下の素数で割り続けると最後には 1 or 素数となる
            # 背理法を考えれば自明 (残された数が √n より上の素数の積であると仮定。これは自明に n を超えるため矛盾)
            for p in candidate_prime_numbers:
                # これ以上調査は無意味
                if num == 1:
                    break
                while num % p == 0:
                    num //= p
                    factorized_dict[p] += 1
            if num != 1:
                factorized_dict[num] = 1
            return factorized_dict

    # import time
    # s = time.time()

    n = ii()
    L = lmi()
    a_max = max(L)
    # O(AlglgA) で前処理
    eratos = Eratos(a_max)
    # print('preprocessed (Eratostenes)')
    # print(time.time() - s)

    # O(n) * O(√A) で LCM を素因数分解した形で保存する
    lcm_dict = eratos.prime_factorize(L[0])
    for i in range(1, n):
        d = eratos.prime_factorize(L[i])
        for k, v in d.items():
            lcm_dict[k] = max(lcm_dict[k], v)
    # print('prime factorized')
    # print(time.time() - s)
    
    # O(√A)
    lcm_mod = 1
    for k, v in lcm_dict.items():
        lcm_mod = (lcm_mod * pow(k, v, mod)) % mod
    
    # O(n)
    ans = 0
    for a in L:
        # a^p-2 ≡ 1/a
        inverse_element = pow(a, mod-2, mod)
        ans = (ans + lcm_mod * inverse_element) % mod
    
    print(ans)
    # print(time.time() - s)

    

if __name__ == "__main__":
    main()

Buggy python submission (verdict: Time Limit Exceeded):
#!/usr/bin/env python3

import sys
# import math
# import re                                    # re.compile(pattern) => ptn obj; p.search(s), p.match(s), p.finditer(s) => match obj; p.sub(after, s)
# 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 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 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                # product(iter, repeat=n)
# from itertools import accumulate             # accumulate(iter[, f])
# 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
# 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 copy import deepcopy                    # to copy multi-dimentional matrix without reference
from fractions import gcd                    # for Python 3.4


def main():
    mod = 1000000007                  # 10^9+7
    inf = float('inf')
    sys.setrecursionlimit(10**6)    # 1000 -> 1000000
    def input(): return sys.stdin.readline().rstrip()
    def ii():  return int(input())
    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 lcm(a, b): return a * b // gcd(a, b)
    
    n = ii()
    L = lmi()
    total_lcm = L[0]
    for i in range(1, n):
        total_lcm = lcm(total_lcm, L[i]) #% mod
    ans = 0
    for elm in L:
        # inverse_element = pow(elm, mod-2, mod)
        # ans = (ans + total_lcm * inverse_element) % mod
        ans = (ans + total_lcm // elm) % mod
    print(ans)
    

if __name__ == "__main__":
    main()

Fixed submission:
#!/usr/bin/env python3

import sys
import math
from collections import defaultdict


def main():
    mod = 1000000007                  # 10^9+7
    sys.setrecursionlimit(10**6)    # 1000 -> 1000000
    def input(): return sys.stdin.readline().rstrip()
    def ii():  return int(input())
    def lmi(): return list(map(int, input().split()))

    class Eratos:
        def __init__(self, num):
            self.table_max = num
            # self.table[i] は i が素数かどうかを示す (bool)
            self.table = [False if i == 0 or i == 1 else True for i in range(num+1)]
            for i in range(2, int(math.sqrt(num)) + 1):
                if self.table[i]:
                    for j in range(i ** 2, num + 1, i):    # i**2 からスタートすることで定数倍高速化できる
                        self.table[j] = False
            # self.table_max 以下の素数を列挙したリスト
            self.prime_numbers = [2] if self.table_max >= 2 else []
            for i in range(3, self.table_max + 1, 2):
                if self.table[i]:
                    self.prime_numbers.append(i)
        
        def is_prime(self, num):
            if num > self.table_max:
                raise ValueError('Eratos.is_prime(): exceed table_max({}). got {}'.format(self.table_max, num))
            return self.table[num]
        
        def prime_factorize(self, num):
            if num > self.table_max:
                raise ValueError('Eratos.prime_factorize(): exceed table_max({}). got {}'.format(self.table_max, num))
            factorized_dict = defaultdict(int)    # 素因数分解の結果を記録する辞書
            candidate_prime_numbers = [i for i in range(2, int(math.sqrt(num)) + 1) if self.is_prime(i)]
            # n について、√n 以下の素数で割り続けると最後には 1 or 素数となる
            # 背理法を考えれば自明 (残された数が √n より上の素数の積であると仮定。これは自明に n を超えるため矛盾)
            for p in candidate_prime_numbers:
                # これ以上調査は無意味
                if num == 1:
                    break
                while num % p == 0:
                    num //= p
                    factorized_dict[p] += 1
            if num != 1:
                factorized_dict[num] = 1
            return factorized_dict

    # import time
    # s = time.time()

    n = ii()
    L = lmi()
    a_max = max(L)
    # O(AlglgA) で前処理
    eratos = Eratos(a_max)
    # print('preprocessed (Eratostenes)')
    # print(time.time() - s)

    # O(n) * O(√A) で LCM を素因数分解した形で保存する
    lcm_dict = eratos.prime_factorize(L[0])
    for i in range(1, n):
        d = eratos.prime_factorize(L[i])
        for k, v in d.items():
            lcm_dict[k] = max(lcm_dict[k], v)
    # print('prime factorized')
    # print(time.time() - s)
    
    # O(√A)
    lcm_mod = 1
    for k, v in lcm_dict.items():
        lcm_mod = (lcm_mod * pow(k, v, mod)) % mod
    
    # O(n)
    ans = 0
    for a in L:
        # a^p-2 ≡ 1/a
        inverse_element = pow(a, mod-2, mod)
        ans = (ans + lcm_mod * inverse_element) % mod
    
    print(ans)
    # print(time.time() - s)

    

if __name__ == "__main__":
    main()

Buggy python submission (verdict: Time Limit Exceeded):
#!/usr/bin/env python3

import sys
# import math
# import re                                    # re.compile(pattern) => ptn obj; p.search(s), p.match(s), p.finditer(s) => match obj; p.sub(after, s)
# 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 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 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                # product(iter, repeat=n)
# from itertools import accumulate             # accumulate(iter[, f])
# 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
# 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 copy import deepcopy                    # to copy multi-dimentional matrix without reference
from fractions import gcd                    # for Python 3.4


def main():
    mod = 1000000007                  # 10^9+7
    inf = float('inf')
    sys.setrecursionlimit(10**6)    # 1000 -> 1000000
    def input(): return sys.stdin.readline().rstrip()
    def ii():  return int(input())
    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 lcm(a, b): return a * b // gcd(a, b)
    
    n = ii()
    L = lmi()
    total_lcm = L[0]
    for i in range(1, n):
        total_lcm = lcm(total_lcm, L[i]) #% mod
    ans = 0
    for elm in L:
        # inverse_element = pow(elm, mod-2, mod)
        # ans = (ans + total_lcm * inverse_element) % mod
        ans = (ans + total_lcm // elm) % mod
    print(ans)
    

if __name__ == "__main__":
    main()

Fixed submission:
#!/usr/bin/env python3

import sys
import math
from collections import defaultdict


def main():
    mod = 1000000007                  # 10^9+7
    sys.setrecursionlimit(10**6)    # 1000 -> 1000000
    def input(): return sys.stdin.readline().rstrip()
    def ii():  return int(input())
    def lmi(): return list(map(int, input().split()))

    class Eratos:
        def __init__(self, num):
            self.table_max = num
            # self.table[i] は i が素数かどうかを示す (bool)
            self.table = [False if i == 0 or i == 1 else True for i in range(num+1)]
            for i in range(2, int(math.sqrt(num)) + 1):
                if self.table[i]:
                    for j in range(i ** 2, num + 1, i):    # i**2 からスタートすることで定数倍高速化できる
                        self.table[j] = False
            # self.table_max 以下の素数を列挙したリスト
            self.prime_numbers = [2] if self.table_max >= 2 else []
            for i in range(3, self.table_max + 1, 2):
                if self.table[i]:
                    self.prime_numbers.append(i)
        
        def is_prime(self, num):
            if num > self.table_max:
                raise ValueError('Eratos.is_prime(): exceed table_max({}). got {}'.format(self.table_max, num))
            return self.table[num]
        
        def prime_factorize(self, num):
            if num > self.table_max:
                raise ValueError('Eratos.prime_factorize(): exceed table_max({}). got {}'.format(self.table_max, num))
            factorized_dict = defaultdict(int)    # 素因数分解の結果を記録する辞書
            candidate_prime_numbers = [i for i in range(2, int(math.sqrt(num)) + 1) if self.is_prime(i)]
            # n について、√n 以下の素数で割り続けると最後には 1 or 素数となる
            # 背理法を考えれば自明 (残された数が √n より上の素数の積であると仮定。これは自明に n を超えるため矛盾)
            for p in candidate_prime_numbers:
                # これ以上調査は無意味
                if num == 1:
                    break
                while num % p == 0:
                    num //= p
                    factorized_dict[p] += 1
            if num != 1:
                factorized_dict[num] = 1
            return factorized_dict

    # import time
    # s = time.time()

    n = ii()
    L = lmi()
    a_max = max(L)
    # O(AlglgA) で前処理
    eratos = Eratos(a_max)
    # print('preprocessed (Eratostenes)')
    # print(time.time() - s)

    # O(n) * O(√A) で LCM を素因数分解した形で保存する
    lcm_dict = eratos.prime_factorize(L[0])
    for i in range(1, n):
        d = eratos.prime_factorize(L[i])
        for k, v in d.items():
            lcm_dict[k] = max(lcm_dict[k], v)
    # print('prime factorized')
    # print(time.time() - s)
    
    # O(√A)
    lcm_mod = 1
    for k, v in lcm_dict.items():
        lcm_mod = (lcm_mod * pow(k, v, mod)) % mod
    
    # O(n)
    ans = 0
    for a in L:
        # a^p-2 ≡ 1/a
        inverse_element = pow(a, mod-2, mod)
        ans = (ans + lcm_mod * inverse_element) % mod
    
    print(ans)
    # print(time.time() - s)

    

if __name__ == "__main__":
    main()