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: Accepted)

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

int main(){
    long long int n; cin>>n;

    long long int a,b,f,min=10000000000;
    
    // 整数/0をやるとfloatin point exceptionのエラーが出るのでi=1から始める
    // i*i<=nでやってもiはなぜかこの条件をすり抜ける
    for(int i=1;i<=n/i;i++){
        f=1;
        if(n%i==0){
            a=i;
            b=n/a;
            while(b>0){
                if(b/10!=0){
                    f++;
                }
                b/=10;
            }
            if(f<min) min=f;
        }
    } 

    cout<<min<<endl;
}