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: Accepted)

#include <bits/stdc++.h>
#define ll long long
#define FOR(i, start, stop) for (ll i = start; i < stop; i++)
using namespace std;

vector<ll> divisor(ll n) {
  vector<ll> ret;
  for(ll i = 1; i * i <= n; i++) {
    if(n % i == 0) {
      ret.push_back(i);
      if(i * i != n) ret.push_back(n / i);
    }
  }
  sort(begin(ret), end(ret));
  return (ret);
}


int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll N;cin>>N;
    vector<ll> dv = divisor(N);
    ll a,b,al,bl;string as,bs;
    ll ans=0,temp=10e10;
    FOR(i,0,dv.size()){
        a=dv[i];b=N/a;
        if(a>b)break;
        as=to_string(a);al=as.size();
        bs=to_string(b);bl=bs.size();
        if(temp>abs(al-bl)){
            temp=abs(al-bl);
            ans=max(al,bl);
        }
    }
    cout<<ans<<endl;
}