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>
#include <math.h>

long int digit(long int num){
    long int dig=0;
    while (num!=0) {
        dig++;
        num/=10;
    }
    return dig;
}

int main()
{
    int j=0;
    long int N,i,flag,a=1,b=1;
    scanf("%ld",&N);
    flag=N;
    for(i=2;i<=(long int)sqrt(N);i++){
        while (flag%i==0) {
            a<b?(a*=i):(b*=i);
            flag/=i;
        }
        if(flag==1) break;
    }
    if(flag==N){
        flag=digit(flag);
        printf("%ld\n",flag);
    }
    else if(flag==1){
        printf("%ld\n",a<b?digit(b):digit(a));
    }
    else{
        if(a<flag&&b<flag){
            a*=b;
            b=flag;
        }
        else a<b?(a*=flag):(b*=flag);
        printf("%ld\n",a<b?digit(b):digit(a));
    }
    return 0;
}