codeforces#P1372B. Omkar and Last Class of Math
Omkar and Last Class of Math
Description
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?
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \leq t \leq 10$). Description of the test cases follows.
Each test case consists of a single integer $n$ ($2 \leq n \leq 10^{9}$).
For each test case, output two positive integers $a$ and $b$, such that $a + b = n$ and $LCM(a, b)$ is the minimum possible.
Input
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \leq t \leq 10$). Description of the test cases follows.
Each test case consists of a single integer $n$ ($2 \leq n \leq 10^{9}$).
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.
Samples
3
4
6
9
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 $LCM$.