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 n;
double ans[500] = {
    1.931851863,   3.196226652,   4.465702516,   5.736856827,   7.008771618,
    8.281094001,   9.553661876,   10.826387273,  12.099221263,  13.372133234,
    14.645100881,  15.918112755,  17.191159216,  18.464230579,  19.737324552,
    21.010436579,  22.283562884,  23.556700742,  24.829849614,  26.103009331,
    27.376172470,  28.649344705,  29.922522157,  31.195704791,  32.468892176,
    33.742083286,  35.015278382,  36.288475927,  37.561684811,  38.834884498,
    40.108084895,  41.381294791,  42.654501727,  43.927713238,  45.200925764,
    46.474140570,  47.747356174,  49.020572937,  50.293790656,  51.567009896,
    52.840229899,  54.113450832,  55.386672596,  56.659896742,  57.933118948,
    59.206343092,  60.479567706,  61.752792837,  63.026018981,  64.299245351,
    65.572472035,  66.845699913,  68.118927335,  69.392156317,  70.665384427,
    71.938613484,  73.211843038,  74.485073056,  75.758303508,  77.031533765,
    78.304764589,  79.577995177,  80.851227575,  82.124458328,  83.397690099,
    84.670922301,  85.944154237,  87.217386633,  88.490620812,  89.763852176,
    91.037085287,  92.310318986,  93.583551970,  94.856787158,  96.130019598,
    97.403253921,  98.676487403,  99.949721250,  101.222955858, 102.496191017,
    103.769424679, 105.042659371, 106.315894353, 107.589220899, 108.862364035,
    110.135599035, 111.408834417, 112.682069853, 113.955304934, 115.228540692,
    116.501781613, 117.775012222, 119.048248242, 120.321484264, 121.594719203,
    122.867955213, 124.141194021, 125.414427525, 126.687663899};
void work() {
  scanf("%d", &n);
  printf("%.9lf\n", ans[n / 2 - 1]);
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) work();
  return 0;
}