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"bits/stdc++.h"
using namespace std;

typedef long long ll;

ll func(ll a, ll b) {
	ll cnta=1,cntb=1;
	while (1) {
		if ((a / 10) != 0) {
			a /= 10;
			cnta++;
		}
		else {
			break;
		}
	}
	while (1) {
		if ((b / 10) != 0) {
			b /= 10;
			cntb++;
		}
		else {
			break;
		}
	}

	return ((cnta>=cntb)?cnta:cntb);
}

int main() {
	ll N;
	cin >> N;

	ll mi=(ll)1e13;
	//cout << sqrt(N) << endl;
	for (ll i = 1; i < sqrt(N); i++) {
		//cout << i << endl;
		ll m;
		if ((N%i) == 0) {
			//cout << i <<" "<< N/i << endl;
			m = func(i, N / i);
			//cout <<" "<< m << endl;
			mi = min(mi, m);
		}
	}
	cout << mi << endl;
}