Vjudge 4.12 省赛训练 B. Sum of Digits

B. Sum of Digits

time limit per test

2 seconds

memory limit per test

265 megabytes

input

standard input

output

standard output

Having watched the last Harry Potter film, little Gerald also decided to practice magic. He found in his father's magical book a spell that turns any number in the sum of its digits. At the moment Gerald learned that, he came across a number n. How many times can Gerald put a spell on it until the number becomes one-digit?

Input

The first line contains the only integer n (0 ≤ n ≤ 10100000). It is guaranteed that n doesn't contain any leading zeroes.

Output

Print the number of times a number can be replaced by the sum of its digits until it only contains one digit.

Examples

input

Copy

0

output

Copy

0

input

Copy

10

output

Copy

1

input

Copy

991

output

Copy

3

Note

In the first sample the number already is one-digit — Herald can't cast a spell.

The second test contains number 10. After one casting of a spell it becomes 1, and here the process is completed. Thus, Gerald can only cast the spell once.

The third test contains number 991. As one casts a spell the following transformations take place: 991 → 19 → 10 → 1. After three transformations the number becomes one-digit.

emmmm，这道题其实蛮简单

就是自己做的时候看到10的10万次就懵了

gets输入，int记录，在未成一个数字前不断循环。

特判 <10 的数

#include 
#include 
#include 
#include 
using namespace std;
char a[1000002];

int main()
{

	while(gets(a))
	{
		
		int l=strlen(a);
		if(l==1) 
		{
			cout<<"0"<<endl;
			continue;
		}
		long long int n=0;
		for(int i=0;i'0';
		}

		int totol=1;
		while(n>=10)
		{
			int l=0;
			int k=n;
			for(int i=1;;i++)
			{
				if(k/10==0) {l++;break;}
				else 
				{
					k=k/10;
					l++;
				}
			}
			int sum=0;
			int kk=n;
			for(int i=1;i<=l;i++)
			{
				sum+=kk%10;
				kk=kk/10;
			}
			n=sum;
			totol++;
		}
		cout<endl;
	}
	return 0;
}