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: Time Limit Exceeded)

#include <stdio.h>
#include <math.h>
//#define BUFSIZE 100

	int f(long int x, long int y);

	int main() {
		long int n, a, b;
		int min, r;
		scanf("%ld",&n);
		if(n<1 || n>10000000000){
			printf("error\n");
			return 1;
		}
		a = 1;
		b = n;
		double t = n;
		min = f(a,b);
		while(a < sqrt(t)){
			do{
				a++;
			}while(n % a != 0);
			b = n / a;
			r = f(a,b);
			if(r < min) min = r;
		}
		printf("%d",min);
		return 0;
	}
	
	int f(long int x, long int y){
		/*char buf[BUFSIZE];
		if(x >= y) return sprintf(buf,"%ld",x);
		else return sprintf(buf,"%ld",y);*/
		int c = 0;
		if(x >= y){
			while(x>0){
				x = x / 10;
				c++;
			}
		}
		else{
			while(y>0){
				y = y / 10;
				c++;
			}
		}
		return c;
	}