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)

#define _CRT_SECURE_NO_WARNINGS
#define _USE_MATH_DEFINES

#include "bits/stdc++.h"
#include <random>
using namespace std;
#define rep(i,n) for(int i=0;i<(n);++i)
#define rep2(i,a,b) for(int i=(a);i<(b);++i)
#define rrep(i,n) for(int i=(n)-1;i>=0;--i)
#define rrep2(i,a,b) for(int i=(a)-1;i>=b;--i)
#define range(i,a,b,c) for(int i=a;\
                    c>0?i<b:\
                    i>b;\
                    i+=c)
#define chmax(a, b) (a = (a) < (b) ? (b) : (a))
#define chmin(a, b) (a = (a) > (b) ? (b) : (a))
using ll = long long;
using ull = unsigned long long;
using ld = long double;
#define all(a) begin(a),end(a)
#define ifnot(a) if(not (a))
#define dump(x)  cerr << #x << " = " << (x) << endl
#define int long long
#ifdef _MSC_VER
const bool test = false;
#else 
const bool test = false;
#endif

int dx[] = { 1,0,-1,0 };
int dy[] = { 0,1,0,-1 };
const int INF = 1 << 28;
const ll INFL = (ll)1 << 58;
ll mod = (int)1e9 + 7;
const double eps = 1e-10;
typedef long double Real;
// return -1, 0, 1
int sgn(const Real& r) { return (r > eps) - (r < -eps); }
int sgn(const Real& a, const Real &b) { return sgn(a - b); }

//.....................
const int MAX = (int)2e5 + 5;

vector<string> split(const string &str, char sep) {
	vector<string> v;
	stringstream ss(str);
	string buffer;
	while (getline(ss, buffer, sep)) {
		v.push_back(buffer);
	}
	return v;
}

template<class InputIterator>
int sum(InputIterator begin, InputIterator end) {
	return accumulate(begin, end, 0ll);
}

string reverse_str(string s) {
	reverse(all(s));
	return s;
}

int n;

void solve() {
	cin >> n;
	int res = INFL;
	rep2(a, 1, n) {
		if (n < a * a) break;
		if (n % a != 0) continue;
		int b = n / a;
		chmin(res, max(to_string(a).size(), to_string(b).size()));
	}
	cout << res << endl;
}

signed main() {
	srand(time(NULL));
	int T = (int)1e15;
	solve();
	cout << fixed << setprecision(15);
	rep(i, T) {
		char s[MAX];
		if (scanf("%s", s) == EOF) break;
		int n = strlen(s);
		for (int i = n - 1; i > -1; i--) {
			ungetc(s[i], stdin);
		}
		solve();
	}
	return 0;
}