Problem: Today, Osama gave Fadi an integer X, and Fadi was wondering about the minimum possible value of max(a, b) such that LCM(a, b) equals X. Both a and b should be positive integers.

LCM(a, b) is the smallest positive integer that is divisible by both a and b. For example, LCM(6, 8) = 24, LCM(4, 12) = 12, LCM(2, 3) = 6.

Of course, Fadi immediately knew the answer. Can you be just like Fadi and find any such pair?

Input

The first and only line contains an integer X (1 ≤ X ≤ 10^{12}).

Output

Print two positive integers, a and b, such that the value of max(a, b) is minimum possible and LCM(a, b) equals X. If there are several possible such pairs, you can print any.

Examples

Input


2


Output


1 2


Input


6


Output


2 3


Input


4


Output


1 4


Input


1


Output


1 1

Solution (in PYTHON3):

Solution:
import io
import os

from collections import Counter, defaultdict, deque


def primeFactors(n):
    # Returns a Counter of prime factors of n
    # e.g., 12 returns {2: 2, 3: 1}
    assert n >= 1
    factors = Counter()
    while n % 2 == 0:
        factors[2] += 1
        n //= 2
    x = 3
    while x * x <= n:
        while n % x == 0:
            factors[x] += 1
            n //= x
        x += 2
    if n != 1:
        factors[n] += 1
    return factors


def solve(X,):
    factors = sorted([k ** v for k, v in primeFactors(X).items()])
    best = (X, 1)
    for mask in range(1 << len(factors)):
        check1 = 1
        check2 = 1
        for i in range(len(factors)):
            if mask & (1 << i):
                check1 *= factors[i]
            else:
                check2 *= factors[i]

        assert check1 * check2 == X
        if check1 >= check2:
            best = min(best, (check1, check2))
        else:
            best = min(best, (check2, check1))

    return str(best[0]) + " " + str(best[1])


if __name__ == "__main__":
    input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline

    (X,) = [int(x) for x in input().split()]
    ans = solve(X,)
    print(ans)