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.
for _ in range(int(input())):
	n = int(input())
	# if n is even just n//2 and n//2 is answer here lcm = n//2 only
	if n & 1 == 0:
		print(n//2,n//2)
	else:
	# if n is odd we need to find a factor k of n
	# if n is divisible by k then n-k also divisible by k
	# thus lcm of k and n-k is n-k, now to get min n-k we inc k to max
	# i.e max k which divide n (but not equal to n as 0 and k can give any lcm)
	# now k is largest factor of n (but not n)
		factors = set()
		i = 1
		while i*i <= n:
			if n%i == 0:
				factors.add(i)
				factors.add(n//i)
			i += 1
		factors.remove(n)
		a = max(factors)
		b = n-a
		print(a,b)