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 <cstdio>
#include <algorithm>
#include <iostream>
#include <vector>
#include <map>
#include <unordered_map>
#include <cmath>
#include <queue>
#include <set>

using namespace std;
typedef long long ll;
typedef pair<int, int> P;
#define INF int(1e9)
#define REP(i, n) for(int i = 0; i < (int)(n); i++)
#define REPR(i, n) for(int i = n - 1; i >= 0; i--)
#define REPONE(i, n) for(ll i = 1; i <= (int)(n); i++)
#define EACH(i, mp) for(auto i:mp)
#define FOR(i, m, n) for(int i = m;i < n;i++)
int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};

ll N;
int MAX_N = 10;

// 1234 -> 4 3 2 1
vector<int> each_digit(ll S) {
    int n_len = 0, n_dig[MAX_N];
    while (S > 0) {
        n_dig[n_len++] = S % 10;
        S /= 10;
    }
    vector<int> ret(n_dig, n_dig + n_len);
    return ret;
}

void solve() {
    ll ans = 0;
    int n_dig = each_digit(N).size();
    REPONE(i, pow(10, ceil(n_dig / 2))) {
        if (N % i == 0) ans = max(i, N / i);
    }
    vector<int> v = each_digit(ans);
    cout << v.size() << endl;
}

int main() {
    cin >> N;
    solve();
    return 0;
}