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: Accepted)

#include <stdio.h>
#include <math.h>

int findNumOfDigit(long int x);
int max(int x, int y);
int f(long int a, long int b);


int main()
{
    long int n;
    long int b;
    int minF;
    int tempF;

    scanf("%ld", &n);
    minF = findNumOfDigit(n); // dummy
    for (int i = 1; i <= sqrt(n); i++)
    {
	b = n / i;
	if (n == i * b)
	{
	    tempF = f(i, b);
	    if (tempF < minF)
	    {
		minF = tempF;
	    }
	}
    }

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

int findNumOfDigit(long int x)
{
    return log10(x) + 1;
}

int max(int x, int y)
{
    if (x > y)
    {
	return x;
    }
    else
    {
	return y;
    }
}

int f(long int a, long int b)
{
    return max(findNumOfDigit(a), findNumOfDigit(b));
}