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>
using namespace std;
#define MOD 1000000007
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define rep1(i, n) for(int i = 1; i <= (int)(n); i++)
#define show(x) {for(auto i: x){cout << i << " ";} cout<<endl;}
#define showm(m) {for(auto i: m){cout << m.x << " ";} cout<<endl;}
typedef long long ll;
typedef pair<int, int> P;
ll gcd(int x, int y){ return y?gcd(y, x%y):x;}
ll lcm(ll x, ll y){ return (x*y)/gcd(x,y);}

//tag 桁数導出

int main()
{
    ll n;
    cin >> n;
    //Nの公約数の内、sqrt(N)より小さく、かつn/最大の数の桁数が答え
    ll invans = 1;
    for (ll i = 1; i*i <= n; i++)
    {
        if (n%i == 0){
            invans = i;
        }
    }
    
    //cout << n/invans << endl;
    cout << (int)to_string(n/invans).size() << endl; 
}