CF1665A.GCD vs LCM

传统题 时间 2000 ms 内存 256 MiB 3 尝试 1 已通过 1 标签

GCD vs LCM

You are given a positive integer nn. You have to find 44 positive integers a,b,c,da, b, c, d such that

  • a+b+c+d=na + b + c + d = n, and
  • gcd⁡(a,b)=lcm⁡(c,d)\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)\gcd(a, b) denotes the greatest common divisor of aa and bb, and lcm⁡(c,d)\operatorname{lcm}(c, d) denotes the least common multiple of cc and dd.

Input

The input consists of multiple test cases. The first line contains a single integer tt (1≤t≤1041 \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 nn (4≤n≤1094 \le n \le 10^9) — the sum of aa, bb, cc, and dd.

Output

For each test case output 44 positive integers aa, bb, cc, dd such that a+b+c+d=na + b + c + d = n and gcd⁡(a,b)=lcm⁡(c,d)\gcd(a, b) = \operatorname{lcm}(c, d).

Note

In the first test case gcd⁡(1,1)=lcm⁡(1,1)=1\gcd(1, 1) = \operatorname{lcm}(1, 1) = 1, 1+1+1+1=41 + 1 + 1 + 1 = 4.

In the second test case gcd⁡(2,2)=lcm⁡(2,1)=2\gcd(2, 2) = \operatorname{lcm}(2, 1) = 2, 2+2+2+1=72 + 2 + 2 + 1 = 7.

In the third test case gcd⁡(2,2)=lcm⁡(2,2)=2\gcd(2, 2) = \operatorname{lcm}(2, 2) = 2, 2+2+2+2=82 + 2 + 2 + 2 = 8.

In the fourth test case gcd⁡(2,4)=lcm⁡(2,1)=2\gcd(2, 4) = \operatorname{lcm}(2, 1) = 2, 2+4+2+1=92 + 4 + 2 + 1 = 9.

In the fifth test case gcd⁡(3,5)=lcm⁡(1,1)=1\gcd(3, 5) = \operatorname{lcm}(1, 1) = 1, 3+5+1+1=103 + 5 + 1 + 1 = 10.

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

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