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;
using f = double;
const f INFTY = 42 * 10000;
inline void polygon(int m, f theta, f &minx, f &miny, f &maxx, f &maxy) {
  minx = miny = INFTY;
  maxx = maxy = -INFTY;
  const f internal_angle = (((f)m - 2.0) * M_PI) / (f)m;
  const f half_internal_angle = 0.5 * internal_angle;
  const f half_diagonal_len = 0.5 * tan(half_internal_angle);
  for (int i = 0; i < m; ++i) {
    f px, py;
    sincos(theta + i * 2 * M_PI / m, &px, &py);
    px *= half_diagonal_len;
    py *= half_diagonal_len;
    if (px < minx) minx = px;
    if (px > maxx) maxx = px;
    if (py < miny) miny = py;
    if (py > maxy) maxy = py;
  }
};
f sidelen(int m, f theta) {
  f minx, miny, maxx, maxy;
  polygon(m, theta, minx, miny, maxx, maxy);
  return max(maxx - minx, maxy - miny);
}
f analytic(int m) {
  f best = -INFTY;
  for (int i = 0; i < m; ++i) {
    best = max(best, sidelen(m, (M_PI * (double)i) / (double)m));
  }
  return best;
}
int main() {
  srand(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    scanf("%d", &n);
    const int m = 2 * n;
    printf("%f\n", analytic(m));
  }
  return 0;
}