#include<stdio.h>
#include <math.h>
int main (){
long long int n,i,a,digit;
digit=0;
scanf("%lld",&n);
a=sqrt(n);
for(i=a;i>=1;i--){
if (n%i==0) {
a=n/i;
while(a !=0){
a = a / 10;
++digit;
}
printf("%lld",digit);
break;
}
}
return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

You are given an integer N.
For two positive integers A and B, we will define F(A,B) as the larger of the following: the number of digits in the decimal notation of A, and the number of digits in the decimal notation of B.
For example, F(3,11) = 2 since 3 has one digit and 11 has two digits.
Find the minimum value of F(A,B) as (A,B) ranges over all pairs of positive integers such that N = A \times B.

Constraints

* 1 \leq N \leq 10^{10}
* N is an integer.

Input

The input is given from Standard Input in the following format:


N


Output

Print the minimum value of F(A,B) as (A,B) ranges over all pairs of positive integers such that N = A \times B.

Examples

Input

10000


Output

3


Input

1000003


Output

7


Input

9876543210


Output

6

### Response

```cpp
#include<stdio.h>
#include <math.h>
int main (){
long long int n,i,a,digit;
digit=0;
scanf("%lld",&n);
a=sqrt(n);
for(i=a;i>=1;i--){
if (n%i==0) {
a=n/i;
while(a !=0){
a = a / 10;
++digit;
}
printf("%lld",digit);
break;
}
}
return 0;
}

```