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 9.2.1) (judge verdict: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;

int digit(int x)
{
  int digits = 0;
  while(x > 0){
    x /= 10;
    digits++;
  }
  
  return digits;
}

int main()
{
  int n;
  cin >> n;
  
  int fmin = n,b = 0;
  int amin = 0, bmin = 0, fn = 0;
  for(int a = 1; a*a <= n; a++){
    if( (n % a) != 0){
      continue;
    }
    b = n / a;
    
    amin = digit(a);
    bmin = digit(b);
    fn = max(amin,bmin);
    fmin = min(fmin,fn);
    amin = bmin = fn = 0;    
  }//for
  
  cout << fmin << endl;
}