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;
namespace IO {
char buf[1 << 21], *p1 = buf, *p2 = buf, obuf[1 << 23], *O = obuf;
inline int read() {
  int x = 0, sign = 0;
  char s = getchar();
  while (!isdigit(s)) sign |= s == '-', s = getchar();
  while (isdigit(s)) x = (x << 1) + (x << 3) + (s - '0'), s = getchar();
  return sign ? -x : x;
}
void print(int x) {
  if (x > 9) print(x / 10);
  (*O++ = x % 10 + '0');
}
}  // namespace IO
namespace math {
const int mod = 998244353;
const int maxn = 1e6 + 5;
long long ksm(long long a, long long b) {
  long long s = 1, m = a;
  while (b) {
    if (b & 1) s = s * m % mod;
    m = m * m % mod, b >>= 1;
  }
  return s;
}
long long inv(long long x) { return ksm(x, mod - 2); }
long long fc[maxn], ifc[maxn];
void init_m(int n) {
  fc[0] = 1;
  for (int i = 1; i <= n; i++) fc[i] = fc[i - 1] * i % mod;
  ifc[n] = inv(fc[n]);
  for (int i = n - 1; i >= 0; i--) ifc[i] = ifc[i + 1] * (i + 1) % mod;
}
long long C(long long n, long long m) {
  return fc[n] * ifc[m] % mod * ifc[n - m] % mod;
}
}  // namespace math
using namespace IO;
const int N = 2e5 + 5;
int n;
double P = acos(-1);
void solve() {
  cin >> n;
  n *= 2;
  double R = P / n, p = P / n * 2, sum = 0;
  for (int i = 1; i <= n / 2; i++, R += p) sum += sin(R);
  printf("%.9lf\n", cos(P / n / 2) * sum);
}
int main() {
  int t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}