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"

using namespace std;
using ll = long long;
using IP = pair<int, int>;

const ll INF = 1LL << 62;
#define atcoder(int)1e9+7
#define Endl endl
#define mp make_pair
#define all(v) v.begin(),v.end()
#define pb push_back

ll gcd(ll a, ll b) {//最大公約数
	if (a%b == 0)return b;
	else gcd(b, a%b);
}

ll lcm(ll a, ll b) {//最小公倍数
	return (a / gcd(a, b))*b;
}

int keta(ll i) {
	int a = 0;
	while (i) {
		i /= 10;
		a++;
	}
	return a;
}

int main() {

	ll n;
	cin >> n;
	ll ans = 9999999;

	for (int i = 1; i <= sqrt(n); i++) {
		if (n % i == 0) {
			ll a1 = keta(i);
			ll a2 = keta(1LL * n / i);
			ans = min(max(a1,a2), ans);

		}
	}

	cout << ans << endl;

	return 0;
}