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>
using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair<int,int> ii;
typedef vector<int> vi;
typedef vector<ii> vii;

#define f first
#define s second
#define sz(x) (int)x.size()
#define all(x) begin(x), end(x)
#define rall(x) rbegin(x), rend(x)
#define pb push_back

#define FOR(i, a, b) for (ll i = (a); i < (b); i++)
#define F0R(i, a) FOR(i, 0, a)
#define ROF(i, a, b) for (ll i = (b); i >= (a); i--)
#define R0F(i, a) ROF(i, 0, a-1)
#define trav(x, a) for (auto& x: a)

const int MX  = 2e5+5;
const int MOD = 1e9+7;
const ll  INF = 1061109567; // 4557430888798830399LL
const ld  EPS = 1e-9;
const ld  PI  = acos(-1);

ll F(ll A, ll B) {
    return max(log10(A)+1, log10(B)+1);
}

int main() {

    ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);

    ll N; cin >> N;

    ll res = INF;
    FOR(i, 1, N+1) {
        if (i*i > N) break;

        if (N%i == 0) {
            res = min(res, F(i, N/i));
        }
    }

    cout << res << "\n";
}