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++ (GCC 9.2.1) (judge verdict: Runtime Error)

#pragma GCC optimize("O3")
#include <bits/stdc++.h>
using namespace std;

using ll = long long;
using vl = vector<ll>;
using vc = vector<char>;
using vvl = vector<vector<ll>>;
using vvc = vector<vector<char>>;

#define rep(i, n) for(ll i = 0; i < (ll)(n); i++)
#define repd(i,n) for(ll i = n - 1; i >= 0; i--)
#define all(x) x.begin(), x.end()

ll n, k, m, M = 1e9 + 7;

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

cin >> n;

ll f;
for(ll i = 1; i * i < n; i++) {
    if(n % i == 0) f = i;
}

f = n / f;

ll num = 0, prd = 1;
while(prd <= f) {
    prd *= 10;
    num++;
}

cout << num << endl;
}