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)

/**
 * code generated by JHelper
 * More info: https://github.com/AlexeyDmitriev/JHelper
 * @author
 */

#include <iostream>
#include <fstream>

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
#define tr(container, it) \
        for (auto it = container.begin(); it != container.end(); it++)
#define scontains(c, x) ((c).find(x) != (c).end())   //O(log n)
#define contains(c, x) (find((c).begin(),(c).end(),x) != (c).end()) //O(n)
#define pll pair<ll,ll>
#define pii pair<int,int>
#define mll map<ll,ll>
#define intv(x, a, b)((x)>=a && (x)<=b)
#define  rep(i, begin, end) for (__typeof(end) i = (begin) - ((begin) > (end)); i != (end) - ((begin) > (end)); i += 1 - 2 * ((begin) > (end)))
#define  _for(i, end) for (__typeof(end) i = 0; i < (end); i += 1)
#define all(x) (x).begin(),(x).end()
//#define len(array)  (sizeof(array)/sizeof((array)[0]))
#define what_is(x) cerr << #x << " is " << x << endl;
#define error(args...) { string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); stringstream _ss(_s); istream_iterator<string> _it(_ss); err(_it, args); }
#define mod(x, m) ((((x) % (m)) + (m)) % (m))
const double PI = 2 * acos(.0);
const int INF = 0x3f3f3f3f;
const ll LLINF = 1000000000000000005LL;;
const ll MOD = (ll) (1e9) + 7;
void _mod(ll &x){x = mod(x, MOD);}
//const ll MOD = (ll) 998244353 ;
const double EPS = 1e-10;

template<typename A, size_t N, typename T>
void Fill(A (&array)[N], const T &val) {
    std::fill((T *) array, (T *) (array + N), val);
}


//#define int ll

class CDigitsInMultiplication {
public:
void solve(std::istream& cin, std::ostream& cout) {
    ll N;
    cin >> N;
    int mi =  11;
    rep(i,1ll,(ll)sqrt(N)+1){
        if(N % (ll)i == 0)mi = min (mi , (int)log10(N/i+.5)+1);
    }
    cout << mi << endl;

}
};


#undef int
int main() {
	CDigitsInMultiplication solver;
	std::istream& in(std::cin);
	std::ostream& out(std::cout);
	solver.solve(in, out);
	return 0;
}