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: Wrong Answer)

#include <iostream>
using namespace std;

int main()
{
    // 入力
    int64_t N;
    cin >> N;

    // 1から割っていてNを二つに分ける
    // 桁数の最小値を更新していく
    int64_t res = 99;
    for(int64_t a = 1; a <= N/2; a++)
    {
        if(N % a == 0)
        {
            int64_t b = N / a;
            int c = max(to_string(a).length(), to_string(b).length());
            if(c < res)
            {
                res = c;
            }

            // A, B の桁数が同じか、A の方が大きくなったら終了（計算コスト削減）
            if(to_string(a).length() >= to_string(b).length())
            {
                //cout << "a: " << a << " b: " << b << endl;
                break;
            }
        }
    }

    // 出力
    cout << res << endl;
}