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<cstdio>
#include<vector>
#include<algorithm>
#include<math.h>
using namespace std;

typedef long long ll;

ll digitsofnum(ll n){
    int digits=0;
    while(n){
        n/=10;
        digits++;
    }
    return digits;
}

int main(){
    ll n;
    scanf("%lld",&n);
    ll a=sqrt(n);
    ll ans=1e11;
    for(ll i=1;i<=a;i++){
        if(n%i==0){
            ll tmp=n/i;
            ll f=max(digitsofnum(tmp),digitsofnum(i));
            ans=min(ans,f);
        }
    }
    printf("%lld\n",ans);
}