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++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <iostream>
#include <bits/stdc++.h>
using namespace std;
long long N;
stringstream s;
string temp;


int digit(long long x)
{
    int res=0;
    while(x!=0)
    {
        x/=10;
        res++;
    }
    return res;
}
int main(void)
{
    cin>>N;
    int cnt=INT_MAX;
    int len;
    s.clear();
    s<<N;
    s>>temp;
    len=temp.length();
    cnt=min(cnt,max(1,len));
    s.clear();
    int len1;
    int len2;
    for(long long i=int(sqrt(N)); i<N; i++)
    {
        if(N%i==0)
        {
            len1=digit(i);
            len2=digit(N/i);
            cnt=min(cnt,max(len1,len2));
            break;
        }
    }
    cout << cnt <<endl;
    return 0;
}