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 rep(i,n) for(int i = 0; i < (int)(n); i++)
#define repr(i,n) for(int i = (int)(n); i >= 0; i--)
#define all(v) v.begin(),v.end()
typedef long long ll;

int main(){
    ll N;
    cin >> N;
    vector<pair<ll,ll> > f(0);
    for(ll i = 1; i * i <= N; i++){
        if (N % i == 0){
            f.push_back(pair<ll,ll>(i, N / i));
        }
    }
    ll len_f = f.size();
    ll ans = 12;
    rep(i,len_f){
        ll ans_sub;
        if (f[i].second / 10000000000 != 0) ans_sub = 11;
        else if (f[i].second / 1000000000 != 0) ans_sub = 10;
        else if (f[i].second / 100000000 != 0) ans_sub = 9;
        else if (f[i].second / 10000000 != 0) ans_sub = 8;
        else if (f[i].second / 1000000 != 0) ans_sub = 7;
        else if (f[i].second / 100000 != 0) ans_sub = 6;
        else if (f[i].second / 10000 != 0) ans_sub = 5;
        else if (f[i].second / 1000 != 0) ans_sub = 4;
        else if (f[i].second / 100 != 0) ans_sub = 3;
        else if (f[i].second / 10 != 0) ans_sub = 2;
        else ans_sub = 1;
        ans = min(ans, ans_sub);
    }
    cout << ans << endl;
}