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 __future__ import division, print_function
# import threading
# threading.stack_size(2**27)
# import sys
# sys.setrecursionlimit(10**7)
# sys.stdin = open('inpy.txt', 'r')
# sys.stdout = open('outpy.txt', 'w')
from sys import stdin, stdout
import bisect            #c++ upperbound
import math
import heapq
i_m=9223372036854775807
def inin():
    return int(input())
def stin():
    return input()
def spin():
    return map(int,stin().split())
def lin():                           #takes array as input
    return list(map(int,stin().split()))
def matrix(n):
    #matrix input
    return [list(map(int,input().split()))for i in range(n)]

################################################
def count2Dmatrix(i,list):
    return sum(c.count(i) for c in list)

def modinv(n,p):
    return pow(n,p-2,p)

def GCD(x, y): 
    x=abs(x)
    y=abs(y)
    if(min(x,y)==0):
        return max(x,y)
    while(y): 
        x, y = y, x % y 
    return x
def Divisors(n) : 
    l = []  
    for i in range(1, int(math.sqrt(n) + 1)) :
        if (n % i == 0) : 
            if (n // i == i) : 
                l.append(i) 
            else : 
                l.append(i)
                l.append(n//i)
    return l
prime=[]
def SieveOfEratosthenes(n): 
    global prime
    prime = [True for i in range(n+1)] 
    p = 2
    while (p * p <= n): 
        if (prime[p] == True): 
            for i in range(p * p, n+1, p): 
                prime[i] = False
        p += 1
    f=[]
    for p in range(2, n): 
        if prime[p]: 
            f.append(p)
    return f
q=[]       
def dfs(n,d,v,c):
    global q
    v[n]=1
    x=d[n]
    q.append(n)
    j=c
    for i in x:
        if i not in v:
            f=dfs(i,d,v,c+1)
            j=max(j,f)
            # print(f)
    return j
# d = {}
 
"""*******************************************************"""
for _ in range(inin()):
    n = inin()

    a, b, lcm = 0, 0, 0; temp = 10**9+1
    af, bf = 0, 0
    
    if n%2==0:
        print(n//2, n//2)
    else:
        l = sorted(Divisors(n))
        for i in l:
            if i<=n//2:
                a = i
                b = n - i
                lcm = (a * b) // GCD(a, b)
                if lcm < temp:
                    temp = lcm
                    af = a
                    bf = b
            # print(a, b, lcm)
        # print(i)
    # print(l)
        print(af, bf)