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: Accepted)

#include <stdio.h>
#include <string.h>

int cntDigits(long num)
{
    int count = 0;
    
    while (num > 0) {
        num /= 10;
        count ++;
    }
    
    return count;
}

int main(void)
{
    long N;
    long A, B;
    int ans;
    int cur;
    int digitA, digitB;
    
    scanf("%ld", &N);
    
    ans = cntDigits(N);
    
    for (A = 1; A * A <= N; A ++) {
        if (N % A != 0) {
            continue;
        }
        
        B = N / A;
        
        digitA = cntDigits(A);
        digitB = cntDigits(B);
        
        if (digitA >= digitB) {
            cur = digitA;
        } else {
            cur = digitB;
        }
        
        if (cur < ans) {
            ans = cur;
        }
    }
    
    printf("%d\n", ans);
    

    return 0;
}