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

#define REP(i,n) for(i=1;i<=n;i++)

int iin() {
    int i;
    scanf("%d", &i);
    return i;
}

long long llin() {
    long long i;
    scanf("%lld", &i);
    return i;
}

double din() {
    double i;
    scanf("%lf", &i);
    return i;
}

char cin() {
    char c;
    scanf("%c", &c);
    return c;
}

void stin(char *s) {
    scanf("%s", s);
}

void iout(int i) {
    printf("%d\n", i);
}

void llout(long long i) {
    printf("%lld\n", i);
}

void dout(double i) {
    printf("%lf\n", i);
}

void cout(char c) {
    printf("%c\n", c);
}

void stout(char *s) {
    printf("%s\n", s);
}

int dfs(int v, int n, int *visited, int table[n][n]) {
    int all_visited = 1, i;

    REP(i,n) {
        if (!visited[i]) {
            all_visited = 0;
            break;
        }
    }

    if (all_visited) {
        return 1;
    }

    int count = 0;

    REP(i,n) {
        if (!table[v][i]) continue;
        if (visited[i]) continue;

        visited[i] = 1;
        count += dfs(i, n, visited, table);
        visited[i] = 0;
    }

    return count;
}

int max(int a, int b) {
    return a > b ? a : b;
}

int main() {
    long long n = llin(), i, a, b, f, min = 11;
    REP(i,n) {
        if (n % i != 0) continue;
        a = i;
        b = n / a;
        if (a > b) break;
        f = max((int)(log10(a) + 1), (int)(log10(b) + 1));
        if (min > f) {
            min = f;
        }
    }
    iout(min);
    return 0;
}