CF1665A GCD vs LCM

You are given a positive integer $ n $ . You have to find $ 4 $ positive integers $ a, b, c, d $ such that - $ a + b + c + d = n $ , and - $ \gcd(a, b) = \operatorname{lcm}(c, d) $ . If there are several possible answers you can output any of them. It is possible to show that the answer always exists. In this problem $ \gcd(a, b) $ denotes the [greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor) of $ a $ and $ b $ , and $ \operatorname{lcm}(c, d) $ denotes the [least common multiple](https://en.wikipedia.org/wiki/Least_common_multiple) of $ c $ and $ d $ .

The input consists of multiple test cases. The first line contains a single integer $ t $ ( $ 1 \le t \le 10^4 $ ) — the number of test cases. Description of the test cases follows. Each test case contains a single line with integer $ n $ ( $ 4 \le n \le 10^9 $ ) — the sum of $ a $ , $ b $ , $ c $ , and $ d $ .

For each test case output $ 4 $ positive integers $ a $ , $ b $ , $ c $ , $ d $ such that $ a + b + c + d = n $ and $ \gcd(a, b) = \operatorname{lcm}(c, d) $ .

In the first test case $ \gcd(1, 1) = \operatorname{lcm}(1, 1) = 1 $ , $ 1 + 1 + 1 + 1 = 4 $ . In the second test case $ \gcd(2, 2) = \operatorname{lcm}(2, 1) = 2 $ , $ 2 + 2 + 2 + 1 = 7 $ . In the third test case $ \gcd(2, 2) = \operatorname{lcm}(2, 2) = 2 $ , $ 2 + 2 + 2 + 2 = 8 $ . In the fourth test case $ \gcd(2, 4) = \operatorname{lcm}(2, 1) = 2 $ , $ 2 + 4 + 2 + 1 = 9 $ . In the fifth test case $ \gcd(3, 5) = \operatorname{lcm}(1, 1) = 1 $ , $ 3 + 5 + 1 + 1 = 10 $ .