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>
#include <stdlib.h>


int digits( int n );
int max( int a, int b );


int main(){
	
	int N;
	scanf("%d", &N);
	
	int min = max( digits(1), digits(N) );
	int a = 2;
	int b;
	while( a * a <= N ){
		
		if( N % a == 0 ){
			b = N / a;
			int value = max( digits(a), digits(b) );
			
			if( value < min ){
				min = value;
			}
			
		}
		
		a++;		
		
	}
	
	printf("%d\n", min);	
	
	return 0;
	
}


int digits( int n ){
	
	int i = 1;
	int rest = n;
	while( rest >= 10 ){
		rest = rest / 10;
		i++;
	}
	
	return i;	
	
}


int max( int a, int b ){
	
	if( a >= b ){
		return a;
		
	}else{
		return b;
	}
	
}