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 <math.h>

int main(void)
{
  char arr[100];
  unsigned long long int n;

  fgets(arr, sizeof(arr), stdin);
  sscanf(arr, "%lld", &n);
  unsigned long long int ans = n;
  int ans_s=0;
  for (; ans != 0;) {
    ans_s++;
    ans /= 10;
  }


  int a;
  for (a=1; a*a <= n; a++) {
    if (n % a != 0) {
      continue;
    }

    unsigned long long int b = n/a;
    unsigned long long int r;
    int a_r=0;
    int b_r=0;
    int ar=a;
    int br=b;
  for (;br != 0;) {
    b_r++;
    br /= 10;
  }
  for (;ar != 0;) {
    a_r++;
    ar /= 10;
  }
    
    if (a_r > b_r) {
      r = a_r;
    } else {
      r = b_r;
    }
    if (ans_s > r) {

      ans_s = r;
    }
  }



  printf("%d\n", ans_s);

  
  return 0;
}