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.
def examA():
    T = I()
    ans = []
    for _ in range(T):
        N = I()
        cur = [1 for _ in range(N)]
        ans.append(cur)
    for v in ans:
        print(" ".join(map(str,v)))
    return

def examB():
    def factorization(n):
        arr = defaultdict(int)
        temp = n
        for i in range(2, int(-(-n ** 0.5 // 1)) + 1):
            if temp % i == 0:
                cnt = 0
                while temp % i == 0:
                    cnt += 1
                    temp //= i
                arr[i] = cnt
        if temp != 1:
            arr[temp] = 1
        if arr == []:
            arr[n] = 1
        return arr

    def make_divisors(n):
        divisors = []
        for i in range(1, int(n ** 0.5) + 1):
            if n % i == 0:
                divisors.append(i)
                if i != n // i:
                    divisors.append(n // i)
        # divisors.sort()
        return divisors

    T = I()
    ans = []
    for _ in range(T):
        N = I()
        f = make_divisors(N)
        #print(f)
        if len(f)<=2:
            cur = [1,N-1]
        else:
            cur = [N//f[2],N-N//f[2]]
        ans.append(cur)
    for v in ans:
        print(" ".join(map(str,v)))
    return

def examC():
    T = I()
    ans = []
    for _ in range(T):
        N = I()

        cur = 0
        ans.append(cur)
    for v in ans:
        print(v)
    return

def examD():
    ans = 0
    print(ans)
    return

def examE():
    ans = 0
    print(ans)
    return

def examF():
    ans = 0
    print(ans)
    return

from decimal import getcontext,Decimal as dec
import sys,bisect,heapq
from copy import deepcopy
from heapq import heappop,heappush,heapify
from collections import Counter,defaultdict,deque
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
def I(): return int(input())
def LI(): return list(map(int,sys.stdin.readline().split()))
def DI(): return dec(input())
def LDI(): return list(map(dec,sys.stdin.readline().split()))
def LSI(): return list(map(str,sys.stdin.readline().split()))
def LS(): return sys.stdin.readline().split()
def SI(): return sys.stdin.readline().strip()
global mod,mod2,inf,alphabet,_ep
mod = 10**9 + 7
mod2 = 998244353
inf = 10**18
_ep = dec("0.000000000001")
alphabet = [chr(ord('a') + i) for i in range(26)]
alphabet_convert = {chr(ord('a') + i): i for i in range(26)}

getcontext().prec = 28

sys.setrecursionlimit(2*10**5)

if __name__ == '__main__':
    examB()

"""

"""