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>
#include<vector>
#include<algorithm>
#include<queue>
#include<map>
#include<stack>
#include<cmath>
#include<iomanip>
#include<set>
#include<numeric>
#include<sstream>
#include<random>
#include<cassert>
using namespace std;
typedef long long ll;
#define rep(i, n) for (int i = 0; i < n; ++i)
#define rrep(i, st, n) for (int i = st; i < n; ++i)
using pii = pair<int, int>;
const int inf = 1e9 + 7;
int dy[] = {0, 0, -1, 1, -1, 1, -1, 1};
int dx[] = {1, -1, 0, 0, -1, 1, 1, -1};
#define ceil(a, b) a / b + !!(a % b)

int keta(ll a) {
    int ret = 0;
    while (a) {
        a /= 10;
        ret++;
    }
    return ret;
}
void solve(ll a) {
    map<ll, int> mp;
    ll aa = a;
    for (int i = 2; (ll)i * i <= a; ++i) {
        if (aa % i != 0) continue;
        while (aa % i == 0) {
            aa /= i;
            mp[i]++;
        }
    }
    mp[aa]++;
    if (mp.size() == 0) {
        cout << keta(a) << endl;
        return;
    }
    ll x = 1, y = 1;
    /*
    for (auto e : mp) {
        cout << e.first << " " << e.second << endl;
    }
     */
    auto it = mp.rbegin();
    for (auto i = it; i != mp.rend(); ++i) {
        //cout << i->first << " " << i->second << endl;
        ll s = i->first;
        //cout << s << endl;
        int c = i->second;
        //cout << c << endl;
        while (c--) {
            //cout << c << endl;
            if (x > y) {
                y *= s;
            } else {
                x *= s;
            }
        }
    }
    
    cout << x << " " << y << endl;
    cout << keta(max(x, y)) << endl;
}
int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll n; cin >> n;
    solve(n);
}