codeforces 550C

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 most 100 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 exceed 100 digits.

Output

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

Otherwise, print "YES" in the first line and the resulting number after removing digits from number n in the second line. The printed number must be divisible by 8.

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

Examples

input

3454

output

YES
344

input

10

output

YES
0

input

111111

output

NO
题目的大致意思是，输入一个不超过100位的数字，随意去掉这个数字的一些数字，但不改变数字顺序，能不能得到一个能被8整除的数。8*125=1000，所以只要看千位数后面的数能不能被8整除就可以了。
一位数有0和8的可以被整除，两位数和三位数可以被8整除的直接输出。

#include<iostream>
#include<string>
using namespace std;
int main()
{
    string s;
    cin >> s;
    for (unsigned int i = 0;i < s.size();++i)
    {
        if (s[i] == '8'||s[i]=='0')
        {
            cout << "YES" << endl << s[i] << endl;
            return 0;
        }
        for (unsigned int j = i + 1;j < s.size();++j)
        {
            int b = (s[i] - '0') * 10 + s[j] - '0';
            if (b % 8 == 0)
            {
                cout << "YES" << endl << b << endl;
                return 0;
            }
            for (unsigned int k = j + 1;k < s.size();++k)
            {
                int temp = (s[i] - '0') * 100 + (s[j] - '0') * 10 + s[k] - '0';
                if (temp % 8 == 0)
                {
                    cout << "YES" << endl << temp << endl;
                    return 0;
                }
            }
        }
    }
    cout << "NO" << endl;
    return 0;
}