Notice that the number 123456789 is a 9-digit number consisting exactly the numbers from 1 to 9, with no duplication. Double it we will obtain 246913578, which happens to be another 9-digit number consisting exactly the numbers from 1 to 9, only in a different permutation. Check to see the result if we double it again!

Now you are suppose to check if there are more numbers with this property. That is, double a given number with k digits, you are to tell if the resulting number consists of only a permutation of the digits in the original number.

Input Specification:

Each input file contains one test case. Each case contains one positive integer with no more than 20 digits.

Output Specification:

For each test case, first print in a line “Yes” if doubling the input number gives a number that consists of only a permutation of the digits in the original number, or “No” if not. Then in the next line, print the doubled number.

Sample Input:

1234567899

Sample Output:

Yes
2469135798

程序代码：

#include<stdio.h>
#include<string.h>
char a[21]={0};
char b[21]={0};
int count[10]={0};
int main()
{
    scanf("%s",a);
    int i,tmp=0, len = strlen(a)-1;
    for(i=len;i>=0;i--)
    {
        count[a[i]-'0']++;
        b[i]=(tmp+(a[i]-'0')*2)%10+'0';
        tmp =((tmp+(a[i]-'0')*2))/10;
    }
    if(tmp>0)
    {
        printf("No\n"); 
        printf("1%s",b);    
    }
    else
    {
        for(i=len;i>=0;i--)
        {
            count[b[i]-'0']--;
        }
        for(i=0;i<=9;i++)
        {
            if(count[i]!=0)
            {
                printf("No\n%s",b);
                return 0;
            }
        }
        printf("Yes\n%s",b);
    }
    return 0;
}