Problem:
Score : 300 points

Problem Statement

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.

Sample Input 1

10000

Sample Output 1

3

F(A,B) has a minimum value of 3 at (A,B)=(100,100).

Sample Input 2

1000003

Sample Output 2

7

There are two pairs (A,B) that satisfy the condition: (1,1000003) and (1000003,1). For these pairs, F(1,1000003)=F(1000003,1)=7.

Sample Input 3

9876543210

Sample Output 3

6

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int main(void)
{
    int i,j,k=0;
    long long int n,min=1,max,a[100]={},cut=1;
    
    scanf("%lld",&n);
    
    while(1){
        if(cut==1)cut++;
        else if(cut==2)cut++;
        else cut=cut+2;
        if(n%cut==0){
            a[i]=cut;
            if(n==cut){
                a[i]=n;
                break;
            }
            i++;
            n=n/cut;
            cut=1;
        }
        if(cut==100000000){
            a[i]=n;
            break;
        }
    }
    max=a[i];
    for(j=i-1;j>=0;j--){
        if(max>a[j])min=min*a[j];
        else max=max*a[j];
    }
    printf("%lld %lld",max,min);
    if(max<min)max=min;
    while(1){
        if(max<1)break;
        k++;
        max=max/10;
    }
    printf("%d\n",k);
    return 0;
}