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 of $a$ and $b$ , and $\operatorname{lcm}(c, d)$ denotes the 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 $\gcd(3, 5) = \operatorname{lcm}(1, 1) = 1$ , $3 + 5 + 1 + 1 = 10$ .