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

#define max(a,b) (a>b)?(a):(b)

int llintcmp(const void *a, const void *b) {
	long long c = *(long long *)a, d = *(long long *)b;
	if (c>d) return 1;
	else if (c==d) return 0;
	else return -1;
}

int digits(long long a) {
	long long p = a/10LL;
	int d = 1;
	
	while (p) {
		p /= 10LL;
		d++;
	}
	return d;
}

int main(void) {
	int d, flag=0, i, j, k;
	long long N, p, q;
	long long div[10000];
	scanf("%lld", &N);

	div[0] = 1LL;
	j = 1;
	for (p=2;p*p<=N;p++) {
		if (N%p == 0) div[j++] = p;
		if (p*p == N) flag++;
	}
//	for (i=0;i<j;i++) printf("%lld\t", div[i]);
	if (j == 1) {
		d = digits(N);
	} else if (flag) {
		for (k=j;k<2*j-1;k++) {
			div[k] = N/div[2*j-k-2];
		}
		qsort(div,2*j-1,sizeof(long long), llintcmp);
		d = digits(div[j]);
	} else {
		for (k=j;k<2*j;k++) {
			div[k] = N/div[2*j-k-1];
		}
		qsort(div,2*j,sizeof(long long), llintcmp);
		d = digits(max(div[j],div[j+1]));
	}

//	for (i=0;i<=2*j;i++) printf("%lld\t", div[i]);
	printf("%d\n", d);
	return 0;
}