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 $$$d$$$.
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 $$$\gcd(3, 5) = \operatorname{lcm}(1, 1) = 1$$$, $$$3 + 5 + 1 + 1 = 10$$$. 
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 $$$\gcd(a, b) = \operatorname{lcm}(c, d)$$$.
t = int(input())
a = []
for _ in range(t):
    a.append(str(int(input()) - 3) + ' 1 1 1')
for i in a:
    print(i)
t = int(input())
a = []
for _ in range(t):
    a.append(str(int(input()) - 3))
print(' 1 1 1\n'.join(a), end=' 1 1 1\n')
def main():
    for _ in range(int(input())):
        n = int(input())
        print(f'{n - 3} 1 1 1')
if __name__ == '__main__':
    import sys
    input = sys.stdin.readline
    exit(main())