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>
#define REP(i, n) for (int i = 0; i < n; i++)
#define REPR(i, n) for (int i = n; i >= 0; i--)
#define FOR(i, m, n) for (int i = m; i < n; i++)
#define FORR(i, m, n) for (int i = m; i >= n; i--)
#define INF 1e9;
using namespace std;
typedef long long ll;
int cnt(ll num) {
  int digits = 0;
  while (num > 0) {
    num /= 10;
    digits++;
  }
  return digits;
}
int main() {
  ll N;
  cin >> N;
  int ans = cnt(N);
  for (ll A = 1; A * A <= N; ++A) {
    if (N % A != 0) continue;
    ll B = N / A;
    int tmp = max(cnt(A), cnt(B));
    ans = min(ans, tmp);
  }
  cout << ans;
  return 0;
}