Problem:
Score : 500 points

Problem Statement

Given are N positive integers A_1,...,A_N.

Consider positive integers B_1, ..., B_N that satisfy the following condition.

Condition: For any i, j such that 1 \leq i < j \leq N, A_i B_i = A_j B_j holds.

Find the minimum possible value of B_1 + ... + B_N for such B_1,...,B_N.

Since the answer can be enormous, print the sum modulo (10^9 +7).

Constraints

1 \leq N \leq 10^4

1 \leq A_i \leq 10^6

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 ... A_N

Output

Print the minimum possible value of B_1 + ... + B_N for B_1,...,B_N that satisfy the condition, modulo (10^9 +7).

Sample Input 1

3
2 3 4

Sample Output 1

13

Let B_1=6, B_2=4, and B_3=3, and the condition will be satisfied.

Sample Input 2

5
12 12 12 12 12

Sample Output 2

5

We can let all B_i be 1.

Sample Input 3

3
1000000 999999 999998

Sample Output 3

996989508

Print the sum modulo (10^9+7).

Solution in PyPy3 (2.4.0) (judge verdict: Accepted)

# E - Flatten

n = int(input())
a = list(map(int, input().split()))
assert len(a) == n

modulus = 10 ** 9 + 7
binmod = (bin(modulus - 2))[2:]  # modulus - 2 in binary

MAX = 10 ** 6

primefact = [0 for i in range(MAX + 1)]

def sieve():
    # cf. https://www.geeksforgeeks.org/lcm-of-n-numbers-modulo-m/
    for i in range(2, MAX + 1):
        if primefact[i] == 0:
            for j in range(i, MAX + 1, i):
                if primefact[j] == 0:
                    primefact[j] = i

def factorize(x):
    if primefact[2] == 0:
        sieve()

    vec = {}
    while x >= 2:
        d = primefact[x]
        if d not in vec:
            vec[d] = 0
        vec[d] += 1
        x //= d
    return vec

def value(x):
    p = 1
    for i in x:
        p = (p * (i ** x[i])) % modulus
    return p

def inverse(x):
    p = 1
    for b in binmod:
        p = p * p % modulus
        if b == '1':
            p = p * x % modulus
    return p

def div(n, d):
    return n * inverse(d) % modulus

l = {}
for x in a:
    f = factorize(x)
    for k in f:
        l[k] = max(l[k], f[k]) if k in l else f[k]
lcm = value(l)

s = sum(div(lcm, x) for x in a) % modulus
print(s)