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: Wrong Answer)

#include <iostream>
#include <climits>
#include <stack>
#include <queue>
#include <string>
#include <utility>
#include <string.h>
#include <iomanip>
#include <stdlib.h>
#include <algorithm>
using namespace std;

int INF = 1<<29;
double Pi = 3.141592653589;
const int mod = 1000000007;
// memset(a,0,sizeof(a)); →全部０にする

long long int i,j;

long long int solve(long long int a,long long int b,long long int c){
	return max(a,max(b,c))-min(a,min(b,c));
}
long long int de(long long int a){
	for(int k=1;k<=100;k++){
		a=a/10;
		if(a==0){
			return k;
			break;
		}
	}
}



int main(){
	long long int n;
	long long int ans=INF;
	cin>>n;
	for(i=0;i<n;i++){
		if(i*i>n)break;
	}
	i-=1;
	for(j=1;j<=i;j++){
		if(n%j==0){
			ans = min(ans,de(n/j));
		}
	}
	//output
	cout<<ans<<endl;
	return 0;
}