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: Wrong Answer)

#include <iostream>      // cout, endl, cin
#include <string>        // string, to_string, stoi
#include <vector>        // vector
#include <algorithm>     // min, max, swap, sort, reverse, lower_bound, upper_bound
#include <utility>       // pair, make_pair
#include <tuple>         // tuple, make_tuple
#include <cstdint>       // int64_t, int*_t
#include <cstdio>        // printf
#include <map>           // map
#include <queue>         // queue, priority_queue
#include <set>           // set
#include <stack>         // stack
#include <deque>         // deque
#include <unordered_map> // unordered_map
#include <unordered_set> // unordered_set
#include <bitset>        // bitset
#include <cctype>        // isupper, islower, isdigit, toupper, tolower
#include <cmath>

using namespace std;
typedef long long ll;

#define rep(i, n) for (int i = 0; i < (n); i++)
#define rep2(i, a, b) for (int i = (a); i < (b); ++i)
#define all(a) (a).begin(), (a).end()
#define all2(a, b) (a).begin(), (a).begin() + (b)

int count(int N)
{

        int cnt = 0;
        while (N)
        {
                cnt++;
                N /= 10;
        }
        return cnt;
}

int main()
{
        ll N;
        cin >> N;
        int ans = 100000;
        for (ll A = 1; A < ll(pow(N, 0.5)); A++)
        {
                if (N % A != 0)
                        continue;
                ll B = N / A;

                int tmp = max(count(A), count(B));
                ans = min(ans, tmp);
        }

        cout << ans << endl;

        return 0;
}