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;
#define reps(i,s,n) for(int i = s; i < n; i++)
#define rep(i,n) reps(i,0,n)
#define Rreps(i,n,e) for(int i = n - 1; i >= e; --i)
#define Rrep(i,n) Rreps(i,n,0)
#define ALL(a) a.begin(), a.end()
#define fi first
#define se second
#define endl "\n"
typedef long long ll;
typedef vector<ll> vec;
typedef vector<vec> mat;
#define traceV(v) for(auto x:v)cout<<x<<" ";cout<<endl
#define traceM(mp) for(auto x:mp)cout<<x.first<<" "<<x.second<<endl

//const ll MOD = 998244353;
const ll MOD = (1e+9) + 7;
const ll INF = 2e+9 + 10;
typedef pair<ll, ll> P;

int F(ll A, ll B) {
  int cnt = 0;
  int x = max(A,B);
  while(x) {
    x/=10;
    cnt++;
  }
  return cnt;
}

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);

  ll N; cin >> N;
  int res = 11;  // MAX
  reps(A, 1, 1000001) {  // min(A,B)<=100000=sqrt(max(N))
    ll B;
    if(N%A==0)B = N/A;
    else continue;
    res = min(res, F(A,B));
  }

  cout << res << endl;

  return 0;
}