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 9.2.1) (judge verdict: Runtime Error)

#include <bits/stdc++.h>
#include <stdlib.h>
#include <fstream>
#include <iostream>
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
typedef pair<ll,ll> llP;

ll mod(ll a, ll b){
  ll ret=a%b;
  if(ret<0) ret+=b;
  return ret;
}

ll modpow(ll a,ll b,ll c){
  ll res=1;
  while(b>0){
    if(b&1) res=mod(res*a,c);
    a=mod(a*a,c);
    b>>=1;
  }
  return res;
}

ll digitcal(ll x){
  ll n=0;
  while(x>0){
    x/=10;
    n++;
  }
  return n;
}

int main() {
  ll n;
  cin>>n;
  ll ans=100;
  for(ll i=0;i<=sqrt(n);i++){
    if(n%i==0){
      ll temp=max(digitcal(i),digitcal(n/i));
      ans=min(ans,temp);
    }
  }
  cout<<ans<<endl;
  return 0;
}