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)

#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define rep(i, n) for(ll i = 1; i < n; i++)
ll MOD = 1e9 + 7;
int INF = 1 << 30;
ll INFL = 1LL << 60;

int main() {
   ll N;
   cin >> N;
   int mi = N;
   int hei = sqrt(N);
   rep(i,hei+1){
       if(N%i == 0){
       int j = i; 
       int digit1 = 0;
       int digit2 = 0;
       int k = i;
       while(j!=0){
           j /= 10;
           digit1++;
       }
       int q = N/k;
       while(q!=0){
           q /= 10;
           digit2++; 
       }
       //cout << digit1 << digit2<<endl;
       int digit = max(digit1,digit2);
       //cout << digit <<endl;
       mi = min(digit,mi);
       }
       //else cout << "失敗"<<endl;
   }
    cout << mi <<endl;
}