Codeforces 550C Divisibility by Eight【数学思维题】好题！

C. Divisibility by Eight

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a non-negative integer n, its decimal representation consists of at most100 digits and doesn't contain leading zeroes.

Your task is to determine if it is possible in this case to remove some of the digits (possibly not remove any digit at all) so that the result contains at least one digit, forms a non-negative integer, doesn't have leading zeroes and is divisible by 8. After the removing, it is forbidden to rearrange the digits.

If a solution exists, you should print it.

Input

The single line of the input contains a non-negative integer n. The representation of number n doesn't contain any leading zeroes and its length doesn't exceed100 digits.

Output

Print "NO" (without quotes), if there is no such way to remove some digits from numbern.

Otherwise, print "YES" in the first line and the resulting number after removing digits from numbern in the second line. The printed number must be divisible by8.

If there are multiple possible answers, you may print any of them.

Examples

Input

3454

Output

YES344

Input

10

Output

YES0

Input

111111

Output

NO

题目大意：

让你在原序列中找到一个子序列是八的倍数（相对位子不能改变）。

思路：

125*8==1000.

那么很明显，对于从1000~2000只内的部分，比如1008,确实也是8的倍数，但是不如我们直接找8即可。1016也是8的倍数，但是不如我们直接找16即可。

那么我们O（n）枚举长度为1的子序列，判断是否是8的倍数。

接下来O（n^）枚举长度为2的子序列，判断是否是8的倍数。

最后在O(n^3)枚举长度为3的子序列，判断是否是8的倍数。

如果有，那么就是YES，否则就是NO.

注意不要输出带有前导0的结果。

Ac代码：

#include<stdio.h>#include<string.h>#include<iostream>#include<queue>using namespace std;int dp[1000][8];int vis[10];char a[150];void Slove(){    int n=strlen(a);    for(int i=0;i<n;i++)    {        for(int j=i+1;j<n;j++)        {            for(int k=j+1;k<n;k++)            {                int numi=a[i]-'0';                int numj=a[j]-'0';                int numk=a[k]-'0';                if((numi*100+numj*10+numk)%8==0)                {                    printf("YES\n");                    printf("%d\n",numi*100+numj*10+numk);                    return ;                }            }        }    }    for(int i=0;i<n;i++)    {        for(int j=i+1;j<n;j++)        {            int numi=a[i]-'0';            int numj=a[j]-'0';            if((numi*10+numj)%8==0)            {                printf("YES\n");                printf("%d\n",numi*10+numj);                return ;            }        }    }    for(int i=0;i<n;i++)    {        if((a[i]-'0')%8==0)        {            printf("YES\n");            printf("%c\n",a[i]);            return ;        }    }    printf("NO\n");    return ;}int main(){    while(~scanf("%s",a))    {        Slove();    }}