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 <stdlib.h>
#include <math.h>
#include <string.h>
#define ll long long
#define rep(i,n) for(ll i=0;i<(n);i++)
#define max(p,q) ((p)>(q)?(p):(q))
#define min(p,q) ((p)<(q)?(p):(q))
#define chmax(a,b) ((a)=(a)>(b)?(a):(b))
#define chmin(a,b) ((a)=(a)<(b)?(a):(b))
#define abs(p) ((p)>=(0)?(p):(-(p)))
#define MOD 1000000007
ll power(ll a,ll b){ll r=1;rep(i,b){r*=a;}return r;}
#define swap(a,b) do{ll w=(a);(a)=(b);(b)=w;}while(0)

//your code here!
ll F(ll A, ll B){
    ll a=0, b=0;
    while(A){
        A /= 10;
        a++;
    }
    while(B){
        B /= 10;
        b++;
    }
    return max(a,b);
}

int main(void){
    ll N;
    scanf("%lld", &N);
    ll ansmin=1e15;
    for(ll i=1; i*i<=N; i++) {
        if(N%i == 0){
            ll B=N/i;
            chmin(ansmin, F(i, B));
        }
    }
    printf("%lld\n", ansmin);
    return 0;
}