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 <string>
#include <stack>
#include <queue>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <algorithm>
#include <cmath>

#include <cstdio>

using namespace std;
#define rep(i, n) for (int i = 0; i < (n); i++)
#define repc(i, s, n) for (int i = (s); i <= (n); i++)
#define rrep(i, n) for (int i = (n) - 1; i >= 0; i--)
#define rrepc(i, s, n) for (int i = (s); i >= (n); i--)
typedef long long ll;
typedef unsigned long long ull;

int main()
{
	ll N;
	ll div;

	cin >> N;

	for (ll i = 1; i * i <= N; i++) {
		if (N % i == 0) div = N / i;
	}

	cout << (int)log10(div) + 1 << endl;

	return 0;
}