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 rep(i,n) REP(i,0,n)
#define REP(i,s,e) for(int i=(s); i<(int)(e); i++)
#define repr(i, n) REPR(i, n, 0)
#define REPR(i, s, e) for(int i=(int)(s-1); i>=(int)(e); i--)
#define pb push_back
#define all(r) r.begin(),r.end()
#define rall(r) r.rbegin(),r.rend()
#define fi first
#define se second

typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

const ll INF = 1e9;
const ll MOD = 1e9 + 7;
const double EPS = 1e-8;

template<typename T> T chmax(T& a, const T& b){return a = (a > b ? a : b);}
template<typename T> T chmin(T& a, const T& b){return a = (a < b ? a : b);}

ll g(ll x) {
    if(x == 0) return 1;
    int ret = 0;
    while(x > 0) {
        ret++;
        x /= 10;
    }
    return ret;
}
ll f(ll a, ll b) {
    return max(g(a), g(b));
}

int main(){
    ll n;
    cin >> n;
    ll ans = INF;
    REP(i, 1, sqrt(n)+10) {
        if(n % i != 0) continue;
        chmin(ans, f(i, n / i));
    }
    cout << ans << endl;
    return 0;
}