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, m, n) for (int i = (m); i < (n); ++i)
#define REP(i, n) FOR(i, 0, n)
#define dump(x)  cerr << #x << " = " << (x) << endl
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl

using namespace std;

typedef long long LL;

template<typename T>
ostream& operator << (ostream& os, const vector<T>& vec) {
    os << "[ ";
    REP(i, vec.size()) {
        os << vec[i] << " ";
    }
    os << "]";

    return os;
}

int main(){
    cin.tie(0);
    ios_base::sync_with_stdio(false);
    
    long long N;
    cin >> N;

    int r = floor(sqrt(N));

    REP(i, r) {
        int c = r - i;
        if (N % c == 0) {
            cout << floor(log10(N / c) + 1) << endl;
            break;
        }
    }

}