A

A - Captain Flint and Crew Recruitment

Despite his bad reputation, Captain Flint is a friendly person (at least, friendly to animals). Now Captain Flint is searching worthy sailors to join his new crew (solely for peaceful purposes). A sailor is considered as worthy if he can solve Flint's task.

Recently, out of blue Captain Flint has been interested in math and even defined a new class of integers. Let's define a positive integer [Math Processing Error]x as nearly prime if it can be represented as [Math Processing Error]p⋅q, where [Math Processing Error]1<p<q and [Math Processing Error]p and [Math Processing Error]q are prime numbers. For example, integers [Math Processing Error]6 and [Math Processing Error]10 are nearly primes (since [Math Processing Error]2⋅3=6 and [Math Processing Error]2⋅5=10), but integers [Math Processing Error]1, [Math Processing Error]3, [Math Processing Error]4, [Math Processing Error]16, [Math Processing Error]17 or [Math Processing Error]44 are not.

Captain Flint guessed an integer [Math Processing Error]n and asked you: can you represent it as the sum of [Math Processing Error]4 different positive integers where at least [Math Processing Error]3 of them should be nearly prime.

Uncle Bogdan easily solved the task and joined the crew. Can you do the same?

Input

The first line contains a single integer [Math Processing Error]t ([Math Processing Error]1≤t≤1000) — the number of test cases.

Next [Math Processing Error]t lines contain test cases — one per line. The first and only line of each test case contains the single integer [Math Processing Error]n [Math Processing Error](1≤n≤2⋅105) — the number Flint guessed.

Output

For each test case print:

• YES and [Math Processing Error]4 different positive integers such that at least [Math Processing Error]3 of them are nearly prime and their sum is equal to [Math Processing Error]n (if there are multiple answers print any of them);
• NO if there is no way to represent [Math Processing Error]n as the sum of [Math Processing Error]4 different positive integers where at least [Math Processing Error]3 of them are nearly prime.

You can print each character of YES or NO in any case.Example

Input

7
7
23
31
36
44
100
258

Output

NO
NO
YES
14 10 6 1
YES
5 6 10 15
YES
6 7 10 21
YES
2 10 33 55
YES
10 21 221 6
题意： 定义一类正整数，能够被p∗q表示，其中p、q(1<p<q)均为素数，称之为nearly primenearly prime 。现要求判断整数n，是否能被4个不同整数之和表示，
且其中至少三个整数为nearly primenearly prime (是则，输出YES否则输出NO)
n=31=2×7+2×5+2×3+1其中14,10,6为nearly prime
能用6,10,14,n−30表示吗？注意，当n−30=6or10or14时，出现了重复数字，我们可以构造出比{2∗3,2∗5,2∗7,n}稍大的组合，即{2∗3,2∗5,3∗5,n}

#include <cstdio>
using namespace std;
int main() {
    int t; scanf("%d", &t);
    while (t--) {
        int n;
        scanf("%d", &n);
        if (n <= 30) {
            printf("NO
");
            continue;
        }
        int last = n - 30;
        if (last == 6 || last == 10 || last == 14) {
            printf("YES
6 10 15 %d
", n - 31);
        }
        else {
            printf("YES
6 10 14 %d
", n - 30);
        }
    }
    return 0;
}