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: Accepted)

#include<stdio.h>
#include<math.h>
typedef long long ll;

int dim(ll a){
  int count=1;
  while(a>9){
    a=a/10;
    count++;
  }
  return count;
}//桁数を返す

int minrev(int a,int b){
  if(a>b){
    return a;
  }
  else{
    return b;
  }
}

int main(){
  ll N,a,b;
  int min=50000;
  scanf("%lld",&N);
  int num=0;
  for(int i=sqrt(N);i>0;i--){
    a=i;
    b=N/a;
    if(a*b==N){
      num=minrev(dim(a),dim(b));
      if(num<min){
      //printf("%lld %lld\n",a,b);
      //printf("%d %d\n",dim(a),dim(b));
      min=num;
      }
    }
  }
  printf("%d",min);
}