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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
#define F first
#define S second
#define MP make_pair
#define pb push_back
#define endl '\n'

using namespace std;

typedef long long LL;
typedef pair< int, int > P;
typedef pair< LL, LL > LP;
typedef pair< int, P > iP;
typedef pair< P, P > PP;

static const int INF = INT_MAX;
static const LL LINF = LLONG_MAX;
static const int MIN = INT_MIN;
static const LL LMIN = LLONG_MIN;
static const int MOD = 1000000007;

int dx[] = {0, -1, 1, 0};
int dy[] = {-1, 0, 0, 1};

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    
    LL n;
    cin >> n;
    vector< LL > div;

    for( LL i = 1; i * i <= n; ++i ) {
        if( !( n % i ) ) {
            div.pb( i );

        }
    }

    int dig = INF;
    LL d = 10;
    for( int i = 0; i < div.size(); ++i ) {
        int cnt = 0;
        LL tmp = div[i];
        while( tmp > 0 ) {
            tmp /= d;
            ++cnt;
        }
        tmp = n / div[i];

        int cnt2 = 0;
        while( tmp > 0 ) {
            tmp /= d;
            ++cnt2;
        }

        dig = min( dig, max( cnt, cnt2 ) );
    }

    cout << dig << endl;
    return 0;
}