CodeForces 1388A 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 x as nearly prime if it can be represented as p⋅q, where 1<p<q and p and q are prime numbers. For example, integers 6 and 10 are nearly primes (since 2⋅3=6 and 2⋅5=10), but integers 1, 3, 4, 16, 17 or 44 are not.

Captain Flint guessed an integer n and asked you: can you represent it as the sum of 4 different positiveintegers where at least 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 t (1≤t≤1000) — the number of test cases.

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

Output

For each test case print:

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

6 10 14 15都由质数组成，又因为6和10 10和14都差4，不能用来做填平用的数，所以选择15，d=n-a-b-c，d如果和abc相等，那么d减一加给14即可

#define _CRT_SECURE_NO_WARNINGS
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cmath>
#include<map>
using namespace std;
typedef long long ll;
#define Inf 0x3f3f3f3f
#define Maxn 20



int main() {
    int t, n, a, b, c, d;
    scanf("%d", &t);
    while (t--) {
        scanf("%d", &n);
        a = 6;
        b = 10;
        c = 14;
        d = a + b + c;
        d = n - d;
        if (d == a || d == b || d == c) {
            c = 15, d -= 1;
            printf("YES\n");
            printf("%d %d %d %d\n", a, b, c, d);
        }
        else if (d > 0) {
            printf("YES\n");
            printf("%d %d %d %d\n", a, b, c, d);
        }
        else printf("NO\n");
    }
    return 0;

    
}