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<inttypes.h>

int check(long long int,long long int);

int main(){
	long long int i,n,num;
	int digit;
	int min = 12;

	scanf("%I64d",&n);

	i = 1;
	while( i*i <= n ){
		if( n%i == 0 ){
			/*num = n/i;*/
			num = check(n,n/i);
			digit = 0;
			while( num != 0 ){
				num /= 10;
				digit++;
			}

			if( min > digit ) min = digit;
		}
		i++;
	}

	printf("%d\n",min);
		
	return 0;
}

int check( long long int a, long long int b ){
	int la = 0;
	int lb = 0;
	int max;

	while( a > 0 ){
		a /= 10;
		la++;
	}

	while( b > 0 ){
		b /= 10;
		lb++;
	}

	if( la > lb ){
		max = la;
	}else{
		max = lb;
	}

	return max;
}