CF1372B · Omkar and Last Class of Math

• 难度：1300
• 标签：greedy、math、number theory
• 链接：https://codeforces.com/problemset/problem/1372/B
• 时间限制：1 second　内存限制：256 megabytes
• 出现位置：假期自习包/A-数学基础(快速幂/筛法/gcd/组合数取模)

英文原题面

Statement

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 $t$ ($1 \leq t \leq 10$). 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.

样例

样例 1

输入：

3
4
6
9

输出：

2 2
3 3
3 6

样例解释（英文原文）

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 $LCM$.