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 <stdio.h>
#include <math.h>
#ifdef _MSC_VER
#include <conio.h>
#else
#define getch()
#include <climits>
#define _CRT_INT_MAX INT_MAX
#endif

int Digits(long long int A){
	int i=1;
	for(i=1;A>=10;i++){
		A/=10;
	}
	return i;
}

inline int Min(int A,int B){
	return A<=B?A:B;
}

inline long long int Max(int A,int B){
	return A>=B?A:B;
}

int main(){
	long long int N=0;
	int Ans=0;
	scanf(" %lld",&N);
	long long int Sqrt=sqrt(N);
	int Tick=1;
	if(N%2!=0){
		Tick=2;
	}
	for(long long int i=Sqrt;;i+=Tick){
		if(N%i==0){
			Ans=Max(Digits(i),Digits(N/i));
			break;
		}
	}
	printf("%d\n",Ans);
	getch();
}