Codeforces Round 655 (Div. 2)
B
Omkar and Last Class of Math
Codeforces
1300
原题
← AC →
翻译中…

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 $b$.

翻译中…

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.

翻译中…

Can you help Omkar solve his ludicrously challenging math problem?

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Input

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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.

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Each test case consists of a single integer $n$ ($2 \leq n \leq 10^{9}$).

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Output

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For each test case, output two positive integers $a$ and $b$, such that $a + b = n$ and $LCM(a, b)$ is the minimum possible.

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Note

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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 $2 \ 2$.

翻译中…

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 $3 \ 3$.

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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 $LCM$.

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样例

Input #1
3
4
6
9
Output #1
2 2
3 3
3 6
InputCopy OutputCopy
题目信息
题号
1372B
时间限制
1 s
内存限制
256 MB
标签
贪心
数学
数论

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