Problem: The statement of this problem is the same as the statement of problem C2. 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 even integer n (2 ≤ n ≤ 200). 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
2
4
200


Output


1.000000000
2.414213562
127.321336469

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int mul(int a, int b) { return (long long)a * b % 1000000007; }
int power(int a, int b) {
  int ans = 1;
  while (b > 0) {
    if (b & 1) ans = mul(ans, a);
    a = mul(a, a);
    b >>= 1;
  }
  return ans;
}
long long int p[1000100] = {0};
vector<long long int> v[1000000];
void sieve(long long int n) {
  for (long long int i = 2; i * i <= n; i++) {
    if (!p[i])
      for (long long int j = i; j <= n; j += i) {
        if (i != j)
          ;
        p[j] = 1;
      }
  }
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int factorial(long long int n) {
  if (n == 0) return 1;
  return n * factorial(n - 1);
}
struct segmenttree {
  vector<int> st;
  int n;
  void init(int _n) {
    n = _n;
    st.clear();
    st.resize(4 * _n, 0);
  }
  void update(int l, int r, int indup, int val, int node) {
    if (l == r) {
      st[node] += val;
      return;
    } else {
      int mid = (l + r) / 2;
      if (indup >= l && indup <= mid) {
        update(l, mid, indup, val, node * 2 + 1);
      } else {
        update(mid + 1, r, indup, val, node * 2 + 2);
      }
      st[node] = st[2 * node + 1] + st[2 * node + 2];
    }
  }
  int query(int si, int se, int l, int r, int node) {
    if (se < l || si > r || l > r) {
      return 0;
    }
    if (si >= l && se <= r) {
      return st[node];
    }
    int mid = (si + se) / 2;
    int q1 = query(si, mid, l, r, node * 2 + 1);
    int q2 = query(mid + 1, se, l, r, node * 2 + 2);
    return q1 + q2;
  }
  int query(int l, int r) { return query(0, n - 1, l, r, 0); }
  void update(int index, int val) { update(0, n - 1, index, val, 0); }
} tree;
bool chkcow(long long int n, vector<long long int> &a, long long int k,
            long long int mis) {
  long long int sum = 0;
  for (long long int i = 0; i < n; i++) {
    if (a[i] > mis) sum += a[i] - mis;
    if (sum >= k) return true;
  }
  return false;
}
struct cmp {
  bool operator()(const pair<long long int, long long int> &a,
                  const pair<long long int, long long int> &b) const {
    long long int lena = a.second - a.first;
    long long int lenb = b.second - b.first;
    if (lena == lenb) return a.first < b.first;
    return lena > lenb;
  }
};
int32_t main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    n <<= 1;
    double x = 90.0 / n;
    printf("%.8lf\n",
           0.5 / sin(x * 3.1415926535897932384626433832795028 / 180.0));
  }
}