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 factorial as fac
from collections import defaultdict
# from copy import deepcopy
import sys, math
f = None
try:
	f = open('q1.input', 'r')
except IOError:
	f = sys.stdin
if 'xrange' in dir(__builtins__):
	range = xrange
# print(f.readline())
sys.setrecursionlimit(10**2)

	
def print_case_iterable(case_num, iterable):
	print("Case #{}: {}".format(case_num," ".join(map(str,iterable))))

def print_case_number(case_num, iterable):
	print("Case #{}: {}".format(case_num,iterable))

def print_iterable(A):
	print (' '.join(A))

def read_int():
	return int(f.readline().strip())
def read_int_array():
	return [int(x) for x in f.readline().strip().split(" ")]
def rns():
	a =  [x for x in f.readline().split(" ")]
	return int(a[0]), a[1].strip()
def read_string():
	return list(f.readline().strip())
def ri():
	return int(f.readline().strip())
def ria():
	return [int(x) for x in f.readline().strip().split(" ")]
def rns():
	a =  [x for x in f.readline().split(" ")]
	return int(a[0]), a[1].strip()
def rs():
	return list(f.readline().strip())
def bi(x):
	return bin(x)[2:]





from collections import deque
import math
NUMBER = 10**9 + 7
# NUMBER = 998244353
def factorial(n) : 
	M = NUMBER
	f = 1
  
	for i in range(1, n + 1):  
		f = (f * i) % M # Now f never can  
						# exceed 10^9+7  
	return f
def mult(a,b):
	return (a * b) % NUMBER

def minus(a , b):
	return (a - b) % NUMBER

def plus(a , b):
	return (a + b) % NUMBER

def egcd(a, b):
	if a == 0:
		return (b, 0, 1)
	else:
		g, y, x = egcd(b % a, a)
		return (g, x - (b // a) * y, y)

def modinv(a):
	m = NUMBER
	g, x, y = egcd(a, m)
	if g != 1:
		raise Exception('modular inverse does not exist')
	else:
		return x % m
def choose(n,k):
	if n < k:
		assert false
	return  mult(factorial(n), modinv(mult(factorial(k),factorial(n-k)))) % NUMBER
from collections import deque, defaultdict 
import heapq


from types import GeneratorType
def bootstrap(f, stack=[]):
	def wrappedfunc(*args, **kwargs):
		if stack:
			return f(*args, **kwargs)
		else:
			to = f(*args, **kwargs)
			while True:
				if type(to) is GeneratorType:
					stack.append(to)
					to = next(to)
				else:
					stack.pop()
					if not stack:
						break
					to = stack[-1].send(to)
			return to

	return wrappedfunc


def dfs(g, timeIn, timeOut,depths,parents): 
	# assign In-time to node u 
	cnt = 0
	# node, neig_i, parent, depth
	stack = [[1,0,0,0]]
	while stack:
		v,neig_i,parent,depth = stack[-1]
		parents[v] = parent
		depths[v] = depth
		# print (v)
		if neig_i == 0:
			timeIn[v] = cnt    
			cnt+=1
		while neig_i < len(g[v]):
			u = g[v][neig_i]
			if u == parent:
				neig_i+=1
				continue
			stack[-1][1] = neig_i + 1
			stack.append([u,0,v,depth+1])
			break
		if neig_i == len(g[v]):
			stack.pop()
			timeOut[v] = cnt
			cnt += 1
  
# def isAncestor(u: int, v: int, timeIn: list, timeOut: list) -> str: 
# 	return timeIn[u] <= timeIn[v] and timeOut[v] <= timeOut[u] 

cnt = 0
@bootstrap
def dfs(v,adj,timeIn, timeOut,depths,parents,parent=0,depth=0):
	global cnt
	parents[v] = parent
	depths[v] = depth
	timeIn[v] = cnt
	cnt+=1
	for u in adj[v]:
		if u == parent:
			continue
		yield dfs(u,adj,timeIn,timeOut,depths,parents,v,depth+1)
	timeOut[v] = cnt
	cnt+=1
	yield 

def gcd(a,b): 
    if a == 0: 
        return b 
    return gcd(b % a, a) 
  
# Function to return LCM of two numbers 
def lcm(a,b): 
    return (a*b) / gcd(a,b) 
def get_num_2_5(n):
	twos = 0
	fives = 0
	while n>0 and n%2 == 0:
		n//=2
		twos+=1
	while n>0 and n%5 == 0:
		n//=5
		fives+=1
	return (twos,fives)
def shift(a,i,num):
	for _ in range(num):
		a[i],a[i+1],a[i+2] = a[i+2],a[i],a[i+1]

def equal(x,y):
	return abs(x-y) <= 1e-9
# def leq(x,y):
# 	return x-y <= 1e-9
def getAngle(a, b, c):
    ang = math.degrees(math.atan2(c[1]-b[1], c[0]-b[0]) - math.atan2(a[1]-b[1], a[0]-b[0]))
    return ang + 360 if ang < 0 else ang
def getLength(a,b):
	return math.sqrt((a[0]-b[0])**2 + (a[1]-b[1])**2)


from heapq import heapify, heappush, heappop


def solution(n):
	i = 2
	while i * i <= n:
		if n % i == 0:
			return f'{(i-1)*n//i} {n//i}'
		i+=1
	return f'1 {n-1}'







	
def main():
	T = 1
	T = ri()
	for i in range(T):
		n = ri()
		# s = rs()
		# n,m	= ria()
		# a = ria()

		x = solution(n)
		# continue
		if 'xrange' not in dir(__builtins__):
			print(x)
		else:
			print >>output,str(x)# "Case #"+str(i+1)+':',
	if 'xrange' in dir(__builtins__):
		print(output.getvalue())
		output.close()

if 'xrange' in dir(__builtins__):
	import cStringIO
	output = cStringIO.StringIO()
#example usage:
#    for l in res:
#       print >>output, str(len(l)) + ' ' +  ' '.join(l)

if __name__ == '__main__':
	main()
