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 <iostream>
#include <vector>
#include <array>
#include <set>
#include <cmath>
#include <string>
#include <queue> 
#include <list>
#include <stack>
#include <unordered_map>
#include <algorithm> 
#include <numeric>  
#include <functional>
#include <cassert>
#include <bitset>
#include <cctype>  
#include <iomanip> 
#include <limits>  
// std::numeric_limits<int>::max(), min()
// std::setprecision(X)
// std::islower(), std::isupper(), std::tolower, std::toupper
// std::accumulate(m.begin(), m.end(), 0LL);
// std::sort(begin(),end()) // 0, 1, 2, 3 ,4 (default, less);
// std::priority_queue<int, vector<int>, less<int>> que; // 4, 3, 2, 1 (defualt, less) : 
 
#define rep(i, n) for(int i = 0; i < (int)(n);    i++)
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()

template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
typedef long long ll;

ll gcd ( ll a, ll b ) { if ( b == 0 ) return a; else return gcd ( b, a%b ); }
ll lcm ( ll a, ll b ) { return a * b / gcd(a, b); }


int main (void) {
  std::cin.tie(0);  std::ios::sync_with_stdio(false);

  ll N; std::cin >> N;

  auto f = [](ll x, ll y) -> int {
    std::string A = std::to_string(x);
    std::string B = std::to_string(y);
    return std::max( sz(A), sz(B) );
  };

  int ans = 1e5;
  for(ll i=1; i<=std::sqrt(N); i++) {
    if ( N % i == 0 ) ans = std::min(ans, f(i, N/i));
  }
  std::cout << ans << std::endl;

  return 0;
}