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 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>

int max(int a, int b){
  if(a > b) return a;
  else return b;
}

int min(int a, int b){
  if(a < b) return a;
  else return b;
}

typedef long long ll;

ll check_digit(ll a){

  int digit = 1;
  while( (a / 10) > 0 ){
    ++digit;
    a /= 10;
  }

  return digit;

}

int main(){

  ll n;
  scanf("%lld",&n);

  ll ans =100000000000000000;
  for(ll a = 1; a*a < n; ++a){
    if(n % a != 0) continue;
    ll b = n/a;
    int tmp = max(check_digit(a),check_digit(b));
    ans = min(ans,tmp);
  }

  printf("%d",ans);

  return 0;

}