Background

The digital root of a positive integer is found by summing the digits of the integer. If the resulting value is a single digit then that digit is the digital root. If the resulting value contains two or more digits, those digits are summed and the process is repeated. This is continued as long as necessary to obtain a single digit.

For example, consider the positive integer 24. Adding the 2 and the 4 yields a value of 6. Since 6 is a single digit, 6 is the digital root of 24. Now consider the positive integer 39. Adding the 3 and the 9 yields 12. Since 12 is not a single digit, the process must be repeated. Adding the 1 and the 2 yeilds 3, a single digit and also the digital root of 39.

Input

The input file will contain a list of positive integers, one per line. The end of the input will be indicated by an integer value of zero.

Output

For each integer in the input, output its digital root on a separate line of the output.

Example

Input

24390

Output

63
意思是输入一个数，每个数各个数相加，如果小于10结束输出。大于10各位数再相加只到小于10.
用数学方法除9求余，能整除9的输出9

#include<iostream>using namespace std;int main(){	int i,j;	cin>>i;		while(i!=0)	{		j=i%9;		if(j==0)		cout<<9<<endl;		else	    cout<<j<<endl;	    cin>>i;	}		return 0;}

　　这个结果符合要求但是没通过，找到另外一种方法：用字符串来保存数字可以做到每个数字独立，且每个数字范围为0~9，而我一开始的想法就是输入一个数字的字符串数组，我计算所有数字的总和，然后再把这个和sum的每个数字再存入数字的字符串数组，后面的做法就跟前面一样了，而循环的终止条件就是总和sum<10.

#include<stdio.h>int main(){ int i,m; char s[1000]; while(scanf("%s",s)==1&&s[0]!=‘0‘){   for(m=i=0;s[i];i++)     m+=s[i]-‘0‘; printf("%d\n",m%9==0?9:m%9); } return 0;}

这个代码可以通过

Digital Roots