Codeforces Round #655 (Div. 2) B. Omkar and Last Class of Math

B. Omkar and Last Class of Math

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

In Omkar's last class of math, he learned about the least common multiple, or LCMLCM. LCM(a,b)LCM(a,b) is the smallest positive integer xx which is divisible by both aa and bb.

Omkar, having a laudably curious mind, immediately thought of a problem involving the LCMLCM operation: given an integer nn, find positive integers aa and bb such that a+b=na+b=n and LCM(a,b)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 tt (1≤t≤101≤t≤10). Description of the test cases follows.

Each test case consists of a single integer nn (2≤n≤1092≤n≤109).

Output

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

Example

input

Copy

3
4
6
9

output

Copy

2 2
3 3
3 6

Note

For the first test case, the numbers we can choose are 1,31,3 or 2,22,2. LCM(1,3)=3LCM(1,3)=3 and LCM(2,2)=2LCM(2,2)=2, so we output 2 22 2.

For the second test case, the numbers we can choose are 1,51,5, 2,42,4, or 3,33,3. LCM(1,5)=5LCM(1,5)=5, LCM(2,4)=4LCM(2,4)=4, and LCM(3,3)=3LCM(3,3)=3, so we output 3 33 3.

For the third test case, LCM(3,6)=6LCM(3,6)=6. It can be shown that there are no other pairs of numbers which sum to 99 that have a lower LCMLCM.

题意：给你一个n，然后选择两个正整数a和b，满足a+b=n并且LCM(a,b)是最小的。

解题思路：已知：a+b=n

　　　　　　　　LCM(a,b)=a*b/gcd(a,b)

　　　　　　　　要让LCM尽可能小，gcd(a,b)就要尽可能大

　　　　　　　　可以设b=k*a  (k>0)

　　　　　　　　gcd(a,b)=a;

　　　　　　　　我们可以得到LCM(a,b)=k*a

　　　　　　　　所以b=k*a

　　　　　　　　然后讲b=k*a带入a+b=n中

　　　　　　　　a=n/(k+1) (k>0&&k