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 <bits/stdc++.h>
using namespace std;
#define REP(i, n) for (int i = 0; i < (int)(n); i++)
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
typedef long long ll;
typedef pair<int, int> P;
const int INF = 100100100;
const int MOD = (int)1e9 + 7;
const double EPS = 1e-9;
ll F(ll a, ll b) {
  int keta1 = 0, keta2 = 0;
  while (0 < a) {
    keta1++;
    a /= 10;
  }
  while (0 < b) {
    keta2++;
    b /= 10;
  }
  return max(keta1, keta2);
}
int main() {
  ll N;
  cin >> N;
  queue<pair<ll,ll>> q{};
  for (ll i = 1;i<floor(sqrt(N));i++){
    ll j = N / i;
    if(i*j==N){
      q.push(pair<ll, ll>(i, j));
    }
  }
  ll mn = INF;
  while(!q.empty()){
    pair<ll, ll> p = q.front();
    q.pop();
    mn = min(mn, F(p.first, p.second));
  }
  cout << mn << endl;
}