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>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef pair<int,int> P;

#define INF 1<<30
#define LINF 1ll<<60ll
#define MOD 1000000007
#define pb(a) push_back(a)

#define _overload3(_1,_2,_3,name,...) name
#define _rep(i,n) _repi(i,0,n)
#define _repi(i,a,b) for(int i=a, i##_len=(b); i<i##_len; ++i)
#define REP(...) _overload3(__VA_ARGS__,_repi,_rep,)(__VA_ARGS__)

#define REPR(i, n) for(int i = n;i >= 0;--i)
#define REPARR(i, v) for(int i = 0;i < sz(v);++i)

#define all(vec) (vec.begin()),(vec.end())
#define sz(x) ((int)(x).size())
#define bit(n) (1ll<<(n))

template<class T>bool chmax(T &a,const T &b){if(a<b){a=b;return true;}return false;}
template<class T>bool chmin(T &a,const T &b){if(a>b){a=b;return true;}return false;}

#define YES(n) cout<<((n)?"YES":"NO")<<endl
#define Yes(n) cout<<((n)?"Yes":"No")<<endl
#define yes(n) cout<<((n)?"yes":"no")<<endl
#define poss(n) cout<<((n)?"possible":"impossible")<<endl
#define Poss(n) cout<<((n)?"Possible":"Impossible")<<endl

int keta(ll n){
    return (int)(log10(n)+1);
}

int main(){
    ll n;
    cin>>n;
    ll i=sqrt(n);
    while(true){
        if(n%i==0){
            break;
        }
        i--;
    }
    cout<<max(keta(i),keta(n/i))<<endl;
    
    return 0;
}