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;
vector<long long int> divisor(long long int n){
  vector<long long int> di;
  for(long long int i=1; i*i<=n; i++){
    if(n%i==0 && i!=n/i){
      di.push_back(i);
      di.push_back(n/i);
    }
  }
  return di;
}

long long int keta(long long int n){
  long long int cnt=0;
  while(n>0){
    n /= 10;
    cnt++;
  }
  return cnt;
}

int main(){
  long long int ans=99999999;
  long long int n;
  cin >> n;
  vector<long long int> div = divisor(n);
  for(int i=0; i<div.size(); i++){
    ans = min(ans,max(keta(div[i]),keta(n/div[i])));
  }
  cout << ans << endl;
}