Sexy primes are pairs of primes of the form (p, p+6), so-named since "sex" is the Latin word for "six". (Quoted from http://mathworld.wolfram.com/SexyPrimes.html)

Now given an integer, you are supposed to tell if it is a sexy prime.

Input Specification:

Each input file contains one test case. Each case gives a positive integer N (≤10^8).

Output Specification:

For each case, print in a line Yes if N is a sexy prime, then print in the next line the other sexy prime paired with N (if the answer is not unique, output the smaller number). Or if N is not a sexy prime, print No instead, then print in the next line the smallest sexy prime which is larger than N.

Sample Input 1:

47

Sample Output 1:

Yes
41

Sample Input 2:

21

Sample Output 2:

No
23

 题目大意：性感素数是指形如 (p, p+6) 这样的一对素数。给定一个整数，判断其是否为一个性感素数。若N是一个性感素数，则在一行中输出Yes，并在第二行输出与N配对的另一个性感素数（若配对的数不唯一，输出较小的那个）。若N不是性感素数，则在一行中输出No，然后在第二行输出大于N的最小性感素数。
分析：分别判断p和p-6，p和p+6是否同时为素数，如果是，则为性感素数输出Yes。不是的话就从p+1往后找到第一个满足要求的数。

#include<algorithm>
#include <iostream>
#include  <cstdlib>
#include  <cstring>
#include   <string>
#include   <vector>
#include   <cstdio>
#include    <queue>
#include    <stack>
#include    <ctime>
#include    <cmath>
#include      <map>
#include      <set>
#define INF 0xffffffff
#define db1(x) cout<<#x<<"="<<(x)<<endl
#define db2(x,y) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<endl
#define db3(x,y,z) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<endl
#define db4(x,y,z,r) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<", "<<#r<<"="<<(r)<<endl
#define db5(x,y,z,r,w) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<", "<<#r<<"="<<(r)<<", "<<#w<<"="<<(w)<<endl
using namespace std;

int num[1000005]={1,1},prime[100000];
int getprime()
{
    int t=0;
    for(int i=2;i<=1000;++i)
        if(num[i]==0)
            for(int j=i*i;j<=1000000;j+=i)
                num[j]=1;
    for(int i=2;i<=1000000;++i)
        if(num[i]==0)prime[t++]=i;

    return t;
}

bool judge(int a,int t)
{
    if(a==1)return 0;
    for(int i=0;i<t;++i)
        if(a%prime[i]==0&&a!=prime[i])return 0;
    return 1;
}

void getans(int n,int t)
{
    while(n)
    {
        if(judge(n,t)==1)
        {

            if(judge(n-6,t)==1)
            {
                printf("No\n%d\n",n);
                return;
            }
            else if(judge(n+6,t)==1)
            {
                printf("No\n%d\n",n);
                return;
            }
        }
        n++;
    }
    return;
}

int main(void)
{
    #ifdef test
    freopen("in.txt","r",stdin);
    //freopen("in.txt","w",stdout);
    clock_t start=clock();
    #endif //test

    int cnt=getprime();
    int n,ans=0;scanf("%d",&n);

    bool f=judge(n,cnt),f1;
    if(f)
    {
        f1=judge(n-6,cnt);
        if(!f1)
        {
            f1=judge(n+6,cnt);
            if(f1)printf("Yes\n%d\n",n+6);
            else getans(n+1,cnt);
        }
        else printf("Yes\n%d\n",n-6);

    }
    else getans(n+1,cnt);

    #ifdef test
    clockid_t end=clock();
    double endtime=(double)(end-start)/CLOCKS_PER_SEC;
    printf("\n\n\n\n\n");
    cout<<"Total time:"<<endtime<<"s"<<endl;        //s为单位
    cout<<"Total time:"<<endtime*1000<<"ms"<<endl;    //ms为单位
    #endif //test
    return 0;
}