Educational Codeforces Round 87 (Rated for Div. 2) C2. Not So Simple Polygon Embedding（几何）

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

You are given a regular polygon with 2⋅n2⋅n vertices (it's convex and has equal sides and equal angles) and all its sides have length 11 . Let's name it as 2n2n -gon.

Your task is to find the square of the minimum size such that you can embed 2n2n -gon in the square. Embedding 2n2n -gon in the square means that you need to place 2n2n -gon in the square in such way that each point which lies inside or on a border of 2n2n -gon should also lie inside or on a border of the square.

You can rotate 2n2n -gon and/or the square.

Input

The first line contains a single integer TT (1≤T≤2001≤T≤200 ) — the number of test cases.

Next TT lines contain descriptions of test cases — one per line. Each line contains single odd integer nn (3≤n≤1993≤n≤199 ). Don't forget you need to embed 2n2n -gon, not an nn -gon.

Output

Print TT real numbers — one per test case. For each test case, print the minimum length of a side of the square 2n2n -gon can be embedded in. Your answer will be considered correct if its absolute or relative error doesn't exceed 10−610−6 .

Example

Input

Copy

3
3
5
199

Output

Copy

1.931851653
3.196226611
126.687663595

n是奇数的情况其实本质上就是把n是偶数的情况造出的边与多边形的边平行的正方形转个角度。
类似下图：
其中靠外的正方形表示最终的正方形的大致方向（已经转过了一定角度），靠内的那条线代表真正的最终那个正方形的边。

设a为多边形拆分出的每个小等腰三角形的顶角，l为等腰三角形的腰长,x为转过的角度。
有几何关系：l=0.5/sin(a/2),而多边形几何中心到最终正方形的边的距离d=lcos(x)=cos(x)/(2*sin(a/2))。
这个表达式d对x求导会发现导数小于0，因此是单调递减的。只需要根据x的定义域确定最大值带入即可。
能注意到x的范围是0~a/2，因为角度再大一点就和之前重复了。
但这还不够，继续观察发现角度的范围可以缩小到0~4/a，因为0~a/4和a/4~a/2是对称的。所以直接带入a/4到表达式计算即可。

#include <bits/stdc++.h>
#define PI 3.1415926535898
using namespace std;
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int n;
        cin>>n;
        double ang=360.0/(2*n);
        ang/=2.0;
        double ans=0.0;
        ans=0.5/sin(ang/180.0*PI)*cos(0.5*ang/180.0*PI)*2;
        printf("%.9lf\n",ans);
     } 
}