In Omkar's last class of math, he learned about the least common multiple, or LCM. LCM(a, b) is the smallest positive integer x which is divisible by both a and b.

Omkar, having a laudably curious mind, immediately thought of a problem involving the LCM operation: given an integer n, find positive integers a and b such that a + b = n and LCM(a, b) is the minimum value possible.

Can you help Omkar solve his ludicrously challenging math problem?

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10). Description of the test cases follows.

Each test case consists of a single integer n (2 ≤ n ≤ 10^{9}).

Output

For each test case, output two positive integers a and b, such that a + b = n and LCM(a, b) is the minimum possible.

Example

Input


3
4
6
9


Output


2 2
3 3
3 6

Note

For the first test case, the numbers we can choose are 1, 3 or 2, 2. LCM(1, 3) = 3 and LCM(2, 2) = 2, so we output 2 \ 2.

For the second test case, the numbers we can choose are 1, 5, 2, 4, or 3, 3. LCM(1, 5) = 5, LCM(2, 4) = 4, and LCM(3, 3) = 3, so we output 3 \ 3.

For the third test case, LCM(3, 6) = 6. It can be shown that there are no other pairs of numbers which sum to 9 that have a lower LCM.
from sys import stdin


def inp():
    return stdin.buffer.readline().rstrip().decode("utf8")


def itg():
    return int(stdin.buffer.readline())


def mpint():
    return map(int, stdin.buffer.readline().split())


# ############################## import
# https://github.com/cheran-senthil/PyRival/tree/master/pyrival
from collections import Counter


def is_prime(n):
    """returns True if n is prime else False"""
    if n < 5 or n & 1 == 0 or n % 3 == 0:
        return 2 <= n <= 3
    s = ((n - 1) & (1 - n)).bit_length() - 1
    d = n >> s
    for a in [2, 325, 9375, 28178, 450775, 9780504, 1795265022]:
        p = pow(a, d, n)
        if p == 1 or p == n - 1 or a % n == 0:
            continue
        for _ in range(s):
            p = (p * p) % n
            if p == n - 1:
                break
        else:
            return False
    return True


def gcd(x, y):
    """greatest common divisor of x and y"""
    while y:
        x, y = y, x % y
    return x


def memodict(f):
    """memoization decorator for a function taking a single argument"""

    class memodict(dict):
        def __missing__(self, key):
            ret = self[key] = f(key)
            return ret

    return memodict().__getitem__


def pollard_rho(n):
    """returns a random factor of n"""
    if n & 1 == 0:
        return 2
    if n % 3 == 0:
        return 3

    s = ((n - 1) & (1 - n)).bit_length() - 1
    d = n >> s
    for a in [2, 325, 9375, 28178, 450775, 9780504, 1795265022]:
        p = pow(a, d, n)
        if p == 1 or p == n - 1 or a % n == 0:
            continue
        for _ in range(s):
            prev = p
            p = (p * p) % n
            if p == 1:
                return gcd(prev - 1, n)
            if p == n - 1:
                break
        else:
            for i in range(2, n):
                x, y = i, (i * i + 1) % n
                f = gcd(abs(x - y), n)
                while f == 1:
                    x, y = (x * x + 1) % n, (y * y + 1) % n
                    y = (y * y + 1) % n
                    f = gcd(abs(x - y), n)
                if f != n:
                    return f
    return n


@memodict
def prime_factors(n):
    """returns a Counter of the prime factorization of n"""
    if n <= 1:
        return Counter()
    f = pollard_rho(n)
    return Counter([n]) if f == n else prime_factors(f) + prime_factors(n // f)


# ############################## main

def lcm(x, y):
    return x*y // gcd(x, y)

def brute_force(x):
    a = min(range(1, x // 2 + 1), key=lambda i: lcm(i, x - i))
    return a, x - a


DEBUG = 0

for case in range(itg()):
    n = itg()
    if n & 1 ^ 1:  # is even
        print(n >> 1, n >> 1)
    elif is_prime(n):
        print(1, n - 1)
    else:
        min_f = min(prime_factors(n))
        max_f = n // min_f
        a = 1 * max_f
        b = (min_f - 1) * max_f
        print(a, b)
        if DEBUG:
            assert (a, b) == brute_force(n)

# Please check!
