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)

#define _CRT_SECURE_NO_WARNINGS
#include"bits/stdc++.h"
#define INF          1e9
#define INFLL        1e18
#define EPS          1e-9
#define MOD          1000000007
#define REP(i,n)     for(ll i=0,i##_len=(n);i<i##_len;++i)
#define REP1(i,n)    for(ll i=1,i##_len=(n);i<=i##_len;++i)
#define REPR(i,n)    for(ll i=(n)-1;i>=0;--i)
#define REPR1(i,n)   for(ll i=(n);i>0;--i)
#define REPC(i,obj)  for(auto i:obj)
#define ALL(obj)     (obj).begin(),(obj).end()
#define SETP(n)      cout<<fixed<<setprecision(n)
#define VV(T,h,w)    vector<vector<T>>(h,vector<T>(w))
#define VVI(T,h,w,i) vector<vector<T>>(h,vector<T>(w,i))
using namespace std;
using ll = long long;
template<typename T = ll>inline T in() { T ret; cin >> ret; return ret; }

ll digits(ll n)
{
    ll ans = 0;
    while (n > 0) {
        n /= 10;
        ++ans;
    }
    return ans;
}
ll f(ll a, ll b)
{
    return digits(max(a, b));
}
int main()
{
    ll n = in();
    double rt = sqrt(n);
    ll ans = INFLL;
    REP1(i, n) {
        if (i > rt)break;
        if (n % i == 0) {
            ans = min(ans, f(n / i, i));
        }
    }
    cout << ans << endl;
}