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 math import *
from copy import *
from string import *				# alpha = ascii_lowercase
from random import *
from sys import stdin,stdout
from sys import maxsize
from operator import *				# d = sorted(d.items(), key=itemgetter(1))
from itertools import *
from collections import Counter		# d = dict(Counter(l))
import math
import math
import time
from queue import Queue
# PradeepGhosh_2017076

def seive(n):
	l=[True]
	l=l*(n+1)
	for i in range(2,int(sqrt(n))+1):
		if(l[i]==True):
			val=i*i
			while(val<len(l)):
				l[val]=False
				val+=i
	prime=[]
	for i in range(2,len(l)):
		if(l[i]==True):
			prime.append(i)
	return prime

def dp(l,i,n,val,ans):
	if(i>=n or val<0):
		return 1000000
	elif(val==0):
		return 0
	else:
		if(ans[i]==1000000):
			ans[i]=min(min(1+dp(l,i+1,n,val-l[i],ans),1+dp(l,i,n,val-l[i],ans)),dp(l,i+1,n,val,ans))
		return ans[i]
from queue import Queue
def factors(n):
	l=[]
	for i in range(1,int(sqrt(n))+1):
		if(n%i==0):
			if(n//i==i):
				l.append(i)
			else:
				l.append(i)
				l.append(n//i)
	return l
def check1(s,v,x):
	while(True):
		if(v not in s):
			return v
		else:
			v-=1
def check2(s,v,x):
	while(True):
		if(v not in s):
			return v
		else:
			v+=1

def up(n):
	j=n-1
	while(j>0):
		if(n%j==0):
			break
		j-=1
	return j		
def fac(n):
	c=1
	i=1
	while(i<=n):
		c*=i
		i+=1
	return c
def check_happy(n):
	ss=set()
	print(n)
	while(n!=1):
		val=n
		s=0
		print(val)
		while(val!=0):
			s+=(val%10)**2
			val=val//10
		print(val)
		print("hi")
		if(val in ss):
			return False
		else:
			n=val
			ss.add(n)
	return True
def update(a,b):
	c=1
	val=a
	while(a<b):
		print(c,a)
		c+=c
		a+=a
	print(1)
	while(a>b):
		print()
		a-=val
		c-=1
	return c
import random as rand
def gcd(a,b): 
    if a == 0: 
        return b 
    return gcd(b % a, a) 
# nakdna
if __name__ == '__main__':
	l=seive(1000000)
	s=set(l)
	t=int(input())
	for i in range(t):
		n=int(input())
		if(n%2==0):
			print(n//2,n//2)
		else:
			if(n in s):
				print(1,n-1)
			else:
				mini=9999999999999
				a=-1
				b=-1
				for i in l:
					if(n%i==0):
						lcm=(i*(n-i))//gcd(i,n-i)
						if(lcm<mini):
							mini=lcm
							a=i
							b=n-i
						val=n//i
						lcm=(val*(n-val))//gcd(val,n-val)
						if(lcm<mini):
							mini=lcm
							a=val
							b=n-val
				if(a==-1):
					print(1,n-1)
				else:
					print(a,b)