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 5.4.1) (judge verdict: Wrong Answer)


//#include "pch.h"
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
//#include <set>
//#include <map>
//#include <iostream>

#define ll long long
#define fri(n) for(i=0;i<(n);i++)
#define frj(n) for(j=0;j<(n);i++)
#define min(p,q) ((p)<(q)?(p):(q))
#define max(p,q) ((p)>(q)?(p):(q))
#define INF 1000000000000//10^12

int main(void)
{
	//変数の宣言
	ll int n;

	//よく使う変数
	int i,j,l;
	int flag=0;
	ll int ans=0;
	int count=0;
	ll int temp,temp1,temp2;
	int max,min;
	int len;
	int sum=0;


	//データの読み込み
	scanf("%d",&n);
//	scanf_s("%lld",&n);

	//	printf("nは%dです\n", n);
	//	printf("データの読み込み終了\n");
	//実際の処理

	ans=100;

	for(i=1;i<sqrt(n)+1;i++){
		if(n%i==0){
			count=0;
			temp=n/i;
			while(temp!=0){
				temp=temp/10;
				count++;
			}
			ans=min(ans,count);
		}
	}


	//	printf("計算部分終了\n");

	//出力

	printf("%lld",ans);

	//	printf("結果の出力終了\n");


	return 0;
}