Codeforces Round 628 (Div. 2)
A
EhAb AnD gCd
Codeforces
800
原题
←B →
翻译中…

You are given a positive integer $x$. Find any such $2$ positive integers $a$ and $b$ such that $GCD(a,b)+LCM(a,b)=x$.

翻译中…

As a reminder, $GCD(a,b)$ is the greatest integer that divides both $a$ and $b$. Similarly, $LCM(a,b)$ is the smallest integer such that both $a$ and $b$ divide it.

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It's guaranteed that the solution always exists. If there are several such pairs $(a, b)$, you can output any of them.

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Input

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The first line contains a single integer $t$ $(1 \le t \le 100)$  — the number of testcases.

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Each testcase consists of one line containing a single integer, $x$ $(2 \le x \le 10^9)$.

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Output

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For each testcase, output a pair of positive integers $a$ and $b$ ($1 \le a, b \le 10^9)$ such that $GCD(a,b)+LCM(a,b)=x$. It's guaranteed that the solution always exists. If there are several such pairs $(a, b)$, you can output any of them.

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Note

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In the first testcase of the sample, $GCD(1,1)+LCM(1,1)=1+1=2$.

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In the second testcase of the sample, $GCD(6,4)+LCM(6,4)=2+12=14$.

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样例

Input #1
2
2
14
Output #1
1 1
6 4
InputCopy OutputCopy
题目信息
题号
1325A
时间限制
1 s
内存限制
256 MB
标签
构造
贪心
数论

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