Hungry Student Problem (水题)_cf903a hungry student problemjava-优快云博客

本文链接：https://blog.youkuaiyun.com/codeswarrior/article/details/79909149

本题是一道关于数学求解的问题，主要讨论了如何通过不同大小包装的鸡块组合来达到顾客想要的确切数量。通过使用取模运算，该算法能够有效地判断是否可以找到合适的组合。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Hungry Student Problem

Ivan's classes at the university have just finished, and now he wants to go to the local CFK cafe and eat some fried chicken.

CFK sells chicken chunks in small and large portions. A small portion contains 3 chunks; a large one — 7 chunks. Ivan wants to eat exactly x chunks. Now he wonders whether he can buy exactly this amount of chicken.

Formally, Ivan wants to know if he can choose two non-negative integers a and b in such a way that a small portions and b large ones contain exactly x chunks.

Help Ivan to answer this question for several values of x!

Input

The first line contains one integer n (1 ≤ n ≤ 100) — the number of testcases.

The i-th of the following n lines contains one integer x_i (1 ≤ x_i ≤ 100) — the number of chicken chunks Ivan wants to eat.

Output

Print n lines, in i-th line output YES if Ivan can buy exactly x_i chunks. Otherwise, print NO.

Example

Input

2
6
5

Output

YES
NO

Note

In the first example Ivan can buy two small portions.

In the second example Ivan cannot buy exactly 5 chunks, since one small portion is not enough, but two small portions or one large is too much.

如果能凑成x那么x = a*3+b*7, 所以不断取模最后看能否成0就行

code：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int main(){
    int n;
    scanf("%d",&n);
    while(n--){
        int x;
        scanf("%d",&x);
        while(x % 3){
            if(x < 7) break;
            x -= 7;
        }
        if(x % 3 == 0) printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}