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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<math.h>

int F(int A,int B);



int main(){
	
	int i,ans=11,N;
	
	
	scanf("%d",&N);
	
	
	for(i=1;i<=sqrt(N);i++){
	
	if(N%i==0){
		if(F(i,N/i)<ans){
		ans=F(i,N/i);
		printf("%d\n",i);
		}
	}
	}
	
	
	printf("\n%d\n\n",ans);
	
	
	return 0;
}



int F(int A,int B){
	
	int keta,a=1,b=1;
	
	while(A/10!=0){
		A=A/10;
		a++;
	}
	while(B/10!=0){
		B=B/10;
		b++;
	}
	
	if(a>b){
		keta=a;
	}else {
		keta=b;
	}
	
	return keta;
}