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.
import sys
import heapq, functools, collections
import math, random
from collections import Counter, defaultdict

# available on Google, not available on Codeforces
# import numpy as np
# import scipy

import math

def prime_factors(nr):
    i = 2
    factors = []
    while i <= nr:
        if i > math.sqrt(nr):
            i = nr
        if (nr % i) == 0:
            factors.append(i)
            nr = nr / i
        elif i == 2:
            i = 3
        else:
            i = i + 2
    return factors


def solve(k):  # fix inputs here
    console("----- solving ------")

    if k%2 == 0:
        return "{} {}".format(k//2, k//2)

    if k%3 == 0:
        return "{} {}".format(k//3, 2*k//3)

    lst = sorted(list(set(prime_factors(k))))
    
    if len(lst) > 1:
        f = lst[0]
        return "{} {}".format(k//f, (f-1)*k//f)

    return "{} {}".format(1, k-1)


def console(*args):  # the judge will not read these print statement
    print('\033[36m', *args, '\033[0m', file=sys.stderr)
    return

# fast read all
# sys.stdin.readlines()

for case_num in range(int(input())):
    # read line as a string
    # strr = input()

    # read line as an integer
    k = int(input())
    
    # read one line and parse each word as a string
    # lst = input().split()

    # read one line and parse each word as an integer
    # lst = list(map(int,input().split()))

    # read matrix and parse as integers (after reading read nrows)
    # lst = list(map(int,input().split()))
    # nrows = lst[0]  # index containing information, please change
    # grid = []
    # for _ in range(nrows):
    #     grid.append(list(map(int,input().split())))

    res = solve(k)  # please change
    
    # Google - case number required
    # print("Case #{}: {}".format(case_num+1, res))

    # Codeforces - no case number required
    print(res)
