Problem: The statement of this problem is the same as the statement of problem C1. The only difference is that, in problem C1, n is always even, and in C2, n is always odd.

You are given a regular polygon with 2 ⋅ n vertices (it's convex and has equal sides and equal angles) and all its sides have length 1. Let's name it as 2n-gon.

Your task is to find the square of the minimum size such that you can embed 2n-gon in the square. Embedding 2n-gon in the square means that you need to place 2n-gon in the square in such way that each point which lies inside or on a border of 2n-gon should also lie inside or on a border of the square.

You can rotate 2n-gon and/or the square.

Input

The first line contains a single integer T (1 ≤ T ≤ 200) — the number of test cases.

Next T lines contain descriptions of test cases — one per line. Each line contains single odd integer n (3 ≤ n ≤ 199). Don't forget you need to embed 2n-gon, not an n-gon.

Output

Print T real numbers — one per test case. For each test case, print the minimum length of a side of the square 2n-gon can be embedded in. Your answer will be considered correct if its absolute or relative error doesn't exceed 10^{-6}.

Example

Input


3
3
5
199


Output


1.931851653
3.196226611
126.687663595

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MODA = 1000000007;
long long int ncr(long long int, long long int);
long long int power(long long int, long long int);
long long int modi(long long int);
long long int powm(long long int, unsigned int);
long long int gcd(long long int, long long int);
long long int kadane(long long int[], long long int);
bool isPrime(long long int);
bool isPowerTwo(long long int);
bool isSubstring(string, string);
void generateFactors(long long int);
void generatePrimeFactors(long long int);
void facto(long long int);
void SieveOfEratosthenes(long long int);
void mo_dfs(long long int, long long int);
long long int fact[1025] = {};
bool prime[105];
long long int timer;
long long int S[100], T[100], FT[200];
vector<int> ADJ[100];
vector<long long int> factors;
vector<long long int> primefactors;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int tc = 1;
  cin >> tc;
  while (tc--) {
    int n;
    cin >> n;
    cout << fixed << setprecision(10)
         << cos(3.1415926535897932384626433832795 / (4 * n)) /
                sin(3.1415926535897932384626433832795 / (2 * n))
         << "\n";
  }
  return 0;
}
void facto(long long int n) {
  fact[0] = 1;
  for (int i = 1; i <= n; i++) fact[i] = (i * 1ll * fact[i - 1]) % MODA;
}
long long int power(long long int a, long long int b) {
  long long int res = 1;
  while (b) {
    if (b & 1) res = (res * a) % MODA;
    a = (a * a) % MODA;
    b >>= 1;
  }
  return res;
}
long long int modi(long long int a) { return powm(a, MODA - 2); }
long long int powm(long long int x, unsigned int y) {
  if (y == 0) return 1;
  long long int p = power(x, y / 2) % MODA;
  p = (p * p) % MODA;
  return (y % 2 == 0) ? p : (x * p) % MODA;
}
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
bool isPowerTwo(long long int x) { return (x && !(x & (x - 1))); }
long long int kadane(long long int a[], long long int size) {
  long long int max_so_far = INT_MIN, max_ending_here = 0;
  for (long long int i = 0; i < size; i++) {
    max_ending_here = max_ending_here + a[i];
    if (max_so_far < max_ending_here) max_so_far = max_ending_here;
    if (max_ending_here < 0) max_ending_here = 0;
  }
  return max_so_far;
}
void generateFactors(long long int n) {
  factors.clear();
  for (long long int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      factors.push_back(i);
      if (n / i != i) factors.push_back(n / i);
    }
  }
  factors.push_back(1);
  factors.push_back(n);
  sort(factors.begin(), factors.end());
}
void generatePrimeFactors(long long int n) {
  primefactors.clear();
  for (long long int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      primefactors.push_back(i);
      while (n % i == 0) n = n / i;
    }
  }
  if (n != 1) primefactors.push_back(n);
}
void SieveOfEratosthenes(long long int n) {
  memset(prime, true, sizeof(prime));
  for (long long int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (long long int i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
}
long long int ncr(long long int n, long long int r) {
  if (r == 0) return 1;
  fact[0] = 1;
  for (int i = 1; i <= n; i++) fact[i] = fact[i - 1] * i % MODA;
  return (fact[n] * modi(fact[r]) % MODA * modi(fact[n - r]) % MODA) % MODA;
}
bool isPrime(long long int n) {
  if (n < 2) return false;
  for (long long int i = 2; i * i <= n; i++)
    if (n % i == 0) return false;
  return true;
}
void mo_dfs(long long int node, long long int parent) {
  S[node] = timer;
  FT[timer] = node;
  timer++;
  for (long long int child : ADJ[node]) {
    if (child != parent) mo_dfs(child, node);
  }
  T[node] = timer;
  FT[timer] = node;
  timer++;
}
bool isSubstring(string s1, string s2) {
  if (s1.find(s2) != string::npos)
    return true;
  else
    return false;
}