Problem: KM country has N kinds of coins and each coin has its value a_i.

The king of the country, Kita_masa, thought that the current currency system is poor, and he decided to make it beautiful by changing the values of some (possibly no) coins.

A currency system is called beautiful if each coin has an integer value and the (i+1)-th smallest value is divisible by the i-th smallest value for all i (1 \leq i \leq N-1).

For example, the set {1, 5, 10, 50, 100, 500} is considered as a beautiful system, while the set {1, 5, 10, 25, 50, 100} is NOT, because 25 is not divisible by 10.

Since changing the currency system may confuse citizens, the king, Kita_masa, wants to minimize the maximum value of the confusion ratios. Here, the confusion ratio for the change in the i-th coin is defined as |a_i - b_i| / a_i, where a_i and b_i is the value of i-th coin before and after the structure changes, respectively.

Note that Kita_masa can change the value of each existing coin, but he cannot introduce new coins nor eliminate existing coins. After the modification, the values of two or more coins may coincide.



Input

Each dataset contains two lines. The first line contains a single integer, N, and the second line contains N integers, {a_i}.

You may assume the following constraints:

1 \leq N \leq 20

1 \leq a_1 \lt a_2 \lt... \lt a_N \lt 10^5

Output

Output one number that represents the minimum of the maximum value of the confusion ratios. The value may be printed with an arbitrary number of decimal digits, but may not contain an absolute error greater than or equal to 10^{-8}.

Examples

Input

3
6 11 12


Output

0.090909090909


Input

3
6 11 24


Output

0.090909090909


Input

3
6 11 30


Output

0.166666666667

Solution (in CPP):

Solution:
//#include <bits/stdc++.h>

#include <iostream>
#include <algorithm>

#include <bitset>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <stack>
#include <queue>
#include <deque>
#include <cstring>
#include <string>
#include <utility>
#include <array>
#include <complex>
#include <valarray>

#include <cassert>
#include <cmath>
#include <functional>
#include <iomanip>
#include <chrono>
#include <random>
#include <numeric>


using namespace std;
//#define int long long

typedef long long ll;
typedef unsigned long long ull;
//typedef unsigned __int128 HASH;
typedef pair<int,int> pii;
typedef pair<ll, ll> pll;
typedef pair<ull, ull> pullull;
typedef pair<ll,int> plli;
typedef pair<double,int> pdi;
typedef pair<long double, int> pdbi;
typedef pair<int,pii> pipii;
typedef pair<ll,pll> plpll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<vi> vvi;
typedef vector<vvi> vvvi;
typedef vector<pii> vpii;
typedef vector<vector<int>> mat;

#define rep(i,n) for (int i=0;i<(n);i++)
#define rep2(i,a,b) for (int i=(a);i<(b);i++)
#define rrep(i,n) for (int i=(n);i>0;i--)
#define rrep2(i,a,b) for (int i=(a);i>b;i--)
#define pb push_back
#define fi first
#define se second
#define all(a) (a).begin(),(a).end()
#define rall(a) (a).rbegin(),(a).rend()

const ll hmod1 = 999999937;
const ll hmod2 = 1000000000 + 9;
const int INF = 1<<30;
const ll INFLL = 1LL<<62;
const long double EPS = 1e-12;
const ll mod = 1000000000 + 7;
const int dx4[4] = {1, 0, -1, 0};
const int dy4[4] = {0, 1, 0, -1};
const int dx8[8] = {1, 1, 1, 0, 0, -1, -1, -1};
const int dy8[8] = {0, 1, -1, 1, -1, 0, 1, -1};
const long double pi = 3.141592653589793;

#define addm(X, Y) (X) = ((X) + ((Y) % mod) + mod) % mod
#define inside(y, x, h, w) (0 <= (y) && (y) < (h) && 0 <= (x) && (x) < (w)) ? true : false

//debug
#define DEBUG

#define DUMPOUT cout

#ifdef DEBUG
#define dump(...) DUMPOUT<<#__VA_ARGS__<<" :["<<__FUNCTION__<<":"<<__LINE__<<"]"<<endl; DUMPOUT<<"    "; dump_func(__VA_ARGS__)
#else
#define dump(...)
#endif

void dump_func() {DUMPOUT << endl;};

template <class Head, class... Tail> void dump_func(Head&& head, Tail&&... tail) {
    DUMPOUT << head;
    if (sizeof...(Tail) == 0) DUMPOUT << " ";
    else DUMPOUT << ", ";
    dump_func(std::move(tail)...);
}

//ostream
template<typename T> ostream& operator << (ostream& os, vector<T>& vec) {
    os << "["; for (int i = 0; i<vec.size(); i++) os << vec[i] << (i + 1 == vec.size() ? "" : ", "); os << "]";
    return os;
}

template<typename T, typename U> ostream& operator << (ostream& os, pair<T, U>& pair_var) {
    os << "(" << pair_var.first << ", " << pair_var.second << ")";
    return os;
}

template<typename T, typename U> ostream& operator << (ostream& os, map<T, U>& map_var) {
    os << "[";
    for (auto itr = map_var.begin(); itr != map_var.end(); itr++) {
        os << "(" << itr->first << ", " << itr->second << ")"; itr++;  if(itr != map_var.end()) os << ", "; itr--;
    }
    os << "]";
    return os;
}

template<typename T> ostream& operator << (ostream& os, set<T>& set_var) {
    os << "[";
    for (auto itr = set_var.begin(); itr != set_var.end(); itr++) {
        os << *itr; ++itr; if(itr != set_var.end()) os << ", "; itr--;
    }
    os << "]";
    return os;
}

int n;
double a[25];
double dp[25][200000 + 5];
vector<vector<int>> v(200000 + 5);

/*
a[1] ~ a[n-1] を a[0]に合わせると考えると, a[0] < a[i] (i = 1, 2, .. n - 1) なので
ratio = 1 - x / a[i] を計算すればよく, また今 x = a[0] なので
ratio = 1 - a[0] / a[i] < 1
よって ratio の max の min は1未満に抑えられる
そしてa[i] <= 100000であることから調べるべき数字のmaxは, xが十分大きいとして
ratio = x / a[i] - 1 を考えればよく
x / a[i] - 1 < 1 より, x < 2 * a[i] <= 200000
∴ x < 200000
ratioのmaxのminは1未満に抑えられるから調べるべき数字のminは, xが十分小さいとして
ratio = 1 - x / a[i] を考えればよく
1 - x / a[i] < 1 より 0 < x
∴ 0 < x;
以上から調べるべき数字の範囲は 0 < x < 200000
*/

signed main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  cin >> n;
  rep(i, n) cin >> a[i];
  rep2(i, 1, 200000 + 5) {
    for (int j = 1; j * j <= i; j++) {
      if (i % j == 0) {
        v[i].push_back(j);
        if (j != i / j) v[i].push_back(i / j);
      }
    }
  }
  rep(i, n)rep(j, 200000+ 1) dp[i][j] = 1;
  rep(i, n) {
    if (i == 0) {
      rep2(j, 1, 200000 + 1) dp[i][j] = abs(a[i] - j) / a[i];
    }
    else {
      rep2(j, 1, 200000 + 1) {
        double ratio = abs(a[i] - j) / a[i];
        double mn = 1;
        for (auto k : v[j]) {
          mn = min(mn, dp[i - 1][k]);
        }
        if (ratio < mn) dp[i][j] = mn;
        else dp[i][j] = ratio;
      }
    }
  }
  double ans = 1;
  rep2(j, 1, 200000 + 1) ans = min(ans, dp[n - 1][j]);
  cout << fixed << setprecision(20) << ans << endl;

}