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 FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)   FOR(i,0,n)
#define ALL(obj)   (obj).begin(),(obj).end()

#define debug(var)  do{std::cout << __LINE__ << "> " << #var << ": ";view(var);}while(0)
template<typename T> void view(T e){std::cout << e << std::endl;}
template<typename T> void view(const std::vector<T>& v){for(const auto& e : v){ std::cout << e << " "; } std::cout << std::endl;}
template<typename T> void view(const std::vector<std::vector<T> >& vv){ for(const auto& v : vv){ view(v); } }

using namespace std;
using ll = long long;
using vi = vector<int>;
using vl = vector<ll>;
using vvi = vector<vector<int>>;
using vvl = vector<vector<ll>>;

vector<long long> divisor(long long n) {
    vector<long long> res;
    for (long long i = 1; i * i <= n; ++i) {
        if (n % i == 0) {
            res.push_back(i);
            if (i != n / i) res.push_back(n / i);
        }
    }
    // if sort is required
    sort(res.begin(), res.end());
    return res;
}

void Main() {
	ll N; cin >> N;
	auto res = divisor(N);
	int ans = 20;
	for (auto v : res) {
		ll tmp = max(v, N / v);
		ans = min(ans, (int)to_string(tmp).size());
	}
	cout << ans << endl;
}

int main() {
    cin.tie(nullptr);
	ios_base::sync_with_stdio(false);
	cout << fixed << setprecision(15);
	Main();
}