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++ (GCC 9.2.1) (judge verdict: Accepted)

#include "bits/stdc++.h"
using namespace std;
typedef long long ll;
#define _USE_MATH_DEFINES
#include <math.h>
#define NIL = -1;
#define all(x) x.begin(),x.end()
const ll INF = 1e9;
const ll INFL = 1e18;
const ll mod = 1e9 + 7;
 
int digit(ll x) {
    int digits = 0;

    while(x > 0){
        x /= 10;
        digits++;
    }

    return digits;
}
 
ll gcd(long long a,long long b) {
    if (a < b) swap(a,b);
    if (b == 0) return a;
    
    return gcd(b,a%b);
}

bool is_prime(long long N){
    if (N == 1) return false;
    for (long long i = 2;i * i <= N;i++){
        if (N % i == 0) return false;
    }
    return true;
}
 
ll lcm(ll a,ll b){
    return ((a * b == 0)) ? 0 : (a / gcd(a,b) * b);
}
 
double DegreeToRadian(double degree){
    return degree * M_PI / 180.0;
}
 
 
int dx[4] = {0,1,0,-1};
int dy[4] = {1,0,-1,0};
 
 
///////////////////////////////////////////////////////////////////////////////////

int main(){
    long long n;
    cin >> n;
    int ans = digit(n);
    for (long long i = 1LL;i * i <= n;i++){
        if (n % i != 0) continue;
        long long b = n / i;

        int res = max(digit(i), digit(b));

        if (ans > res){
            ans = res;
        }
    }

    cout << ans << endl;
}