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: Runtime Error)

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#define MAX(X, Y) (((X) > (Y))? (X) : (Y))
#define MIN(X, Y) (((X) < (Y))? (X) : (Y))
int compare_int(const void *a, const void *b){
  return *(long int*)b - *(long int*)a; // 大きい順
}
// 公約数
long int gcd(long int a, long int b){
  if(b == 0){
    return a;
  }else{
    return gcd(b, a % b);
  }
}
// 最小公倍数 = a * b / 最大公約数
// 公倍数
long int lcm(long int a, long int b){
  return a * b / gcd(a, b);
}


// min
int main(void){
  long long int n;
  int ans = 0;
  int keta = 0;

  scanf("%lld", &n);
  //約数を求める。その時の桁数を求める。一番桁数が小さくなる組合せを求める。

  long long int k = n;
  while(k != 0){
    k /= 10;
    ans++;
  }
  
  for(long int i  = 0; i * i <= n; i++){
    printf("i = %ld\n", i);
    if(n % i == 0){
      long int j = n / i;
      while(j != 0){
	j /= 10;
	keta++;
      }
      printf("ans = %d, keta = %d\n", ans, keta);
      if(ans > keta){
	ans = keta;
      }
      keta = 0;
    }
  }
  
  printf("%d\n", ans);
  return 0;
}