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: Time Limit Exceeded)

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>

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

int max(int a, int b){
    return a > b ? a : b;
}

int main(void){
    long int N;
    scanf("%ld" ,&N);
    long int A,B;
    int i;
    int min = INT_MAX;
    int digit_A,digit_B,f;
    for(i=1;i<=N;i++){
        if(N % i == 0){
            A = i;
            B = N / A;
        }
        if(A > B) break;
        digit_A = digit(A);
        digit_B = digit(B);
        f = max(digit_A,digit_B);
        if(f < min) min = f;
    }
    printf("%d\n",min);
}