GCD vs LCM
Description
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 $\gcd(a, b) = \operatorname{lcm}(c, d)$.
Input
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 $d$.
Output
For each test case output $4$ positive integers $a$, $b$, $c$, $d$ such that $a + b + c + d = n$ and $\gcd(a, b) = \operatorname{lcm}(c, d)$.
Samples
5
4
7
8
9
10
1 1 1 1
2 2 2 1
2 2 2 2
2 4 2 1
3 5 1 1
Note
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 $\gcd(3, 5) = \operatorname{lcm}(1, 1) = 1$, $3 + 5 + 1 + 1 = 10$.