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 <bits/stdc++.h>
using namespace std;
int main(){
  long N;
  cin >> N;
  
  vector<int> n(1,1);
  for(long i=2; i<=N/i; i++){
    if(N%i!=0) continue;
    N/=i;
    n.push_back(i);
    i--;
  }
  n.push_back(N);
  int end=n.size(), ans=11;
  long a=1;
  for(int i=0; i<end; i++){
    if(i%2==0){
      a*=n.at(i/2);
      n.at(i/2)=1;
    }
    else{
      a*=n.at(end-1-i/2);
      n.at(end-1-i/2)=1;
    }
    long b=1;
    for(int j=0; j<end; j++){
      b*=n.at(j);
    }
    long x=a, y=b;
    int fx=1, fy=1;
    for(int k=0; k<fx; k++){
      x/=10;
      if(x==0) break;
      fx++;
    }
    for(int k=0; k<fy; k++){
      y/=10;
      if(y==0) break;
      fy++;
    } 
    int fz=max(fx,fy);
    ans=min(ans,fz);
  }
  cout << ans << endl;
}