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 F first
# define S second
# define mp make_pair
# define pii pair<int,int>

# define long long long
# define pb push_back
# define sz(a) (int)(a.size())

# define y1     tipa_y1
# define left   tipa_left
# define right  tipa_right

const int Mod = (int)1e9 + 7;
const int MX = 1073741822;
const long MXLL = 4611686018427387903;
const int Sz = 1110111;

using namespace std;

inline void Read_rap () {
  ios_base :: sync_with_stdio(0);
  cin.tie(0);
}
long n;

inline int f (long a) {
  int res = 0;
  while (a) {
    res ++;
    a /= 10;
  }
  return res;
}

inline int f (long a, long b) {
  return max (f(a), f(b));
}

int main()
{
  Read_rap ();
  cin >> n;

  int mn = MX;    
  for (long i = 1;i * i <= n;i ++) {
    if (n % i == 0)
      mn = min (mn, f(i, n / i));
  }
  cout << mn;

  return 0;
}









// Coded by Z...