Educational Codeforces Round 34 A. Hungry Student Problem【枚举】-优快云博客

本文探讨了一个关于购买鸡块的趣味算法问题，通过双重枚举的方法解决如何根据小份和大份的数量来准确购买指定数量的鸡块，提供了一种简单有效的解决方案。

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 xi (1 ≤ xi ≤ 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 xi 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.

【分析】：刚开始以为是拓展欧几里得解不定方程（被骚操作冲昏头脑的我）···后来才发现两个for枚举就行。

【代码】：

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define oo 10000000
const int mod = 1e6;
int l,n,m;
int ans;
int main()
{
        int t,f;
        scanf("%d",&t);
        while(t--)
        {
            f=0;
            scanf("%d",&n);
            for(int i=0;i<105;i++)
            {
                for(int j=0;j<105;j++)
                {
                    if(3*i+7*j==n)
                    {
                        f=1;
                        break;
                    }
                }
                if(f) break;
            }
            //printf("%d\n",f);
            if(f) puts("YES");
            else puts("NO");
        }
    return 0;
}