#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")

#define _CRT_SECURE_NO_WARNINGS

#include <bits/stdc++.h>
using namespace std;

#include <boost/multiprecision/cpp_int.hpp>
using Bint = boost::multiprecision::cpp_int; 


int main() {
	// 係数列を前もって計算しておく．
	int N = 200;
	vector<vector<Bint>> f(N + 1);
	f[0] = vector<Bint>{ 1 };
	for (int n = 1; n <= N; ++n) {
		f[n].resize(n + 1);
		for (int i = 0; i < n; ++i) {
			f[n][i + 1] -= 2 * n * f[n - 1][i];
		}
		for (int i = 1; i < n; ++i) {
			f[n][i - 1] += i * f[n - 1][i];
			f[n][i + 1] += i * f[n - 1][i];
		}
		for (int i = 0; i <= n; ++i) {
			f[n][i] /= n;
		}
	}

	vector<Bint> pow100000(N + 1);
	pow100000[0] = 1;
	for (int n = 0; n < N; ++n) {
		pow100000[n + 1] = pow100000[n] * 100000;
	}

	for (int n = 1; n <= N; ++n) {
		for (int i = 0; i <= n; ++i) {
			f[n][i] *= pow100000[n - i];
		}
	}

	int t;
	scanf("%d", &t);

	// クエリ先読み
	vector<long long> nXjs(t);
	for (int j = 0; j < t; ++j) {
		int n; float x;
		scanf("%d %f", &n, &x);

		long long X_int = (long long)round(x * 100000.f);

		nXjs[j] = ((long long)n << 40) | ((X_int + 100000) << 20) | (long long)j;
	}

	// クエリの処理順をランダムにする．
	mt19937_64 mt(0);
	shuffle(nXjs.begin(), nXjs.end(), mt);

	vector<float> res(t);

	// 答えのメモ
	vector<map<int, float>> sol(201);

	for (auto nXj : nXjs) {
		int j = (int)(nXj & ((1LL << 20) - 1));
		nXj >>= 20;
		int X_int = (int)(nXj & ((1LL << 20) - 1)) - 100000;
		int n = (int)(nXj >> 20);
		
		if (n == 0) {
			res[j] = 1;
			continue;
		}

		// メモした答えの線形補間
		auto nit = sol[n].lower_bound(X_int);
		if (nit != sol[n].begin() && nit != sol[n].end()) {
			auto pit = prev(nit);

			auto [pi, pv] = *pit;
			auto [ni, nv] = *nit;

			// メモした答えの幅 40 以下の区間に挟まれているなら内挿補間する
			if (ni - pi <= 40) {
				float val = pv + (nv - pv) / (ni - pi) * (X_int - pi);
				res[j] = val;
				continue;
			}
		}

		Bint X = X_int;
		Bint X2 = X * X;

		// ここが遅い・・・はずなのだが思ってるよりずっと速い
		Bint Res = 0;
		for (int i = n; i >= 0; i -= 2) {
			Res *= X2;
			Res += f[n][i];
		}
		if (n & 1) Res *= X;

		float val = static_cast<float>(Res / pow100000[n - 1]);
		float x = X_int / 100000.f;
		val *= pow(x * x + 1, -n / 2.f) / 100000.f;

		res[j] = val;

		sol[n][X_int] = val;
	}

	for (int j = 0; j < t; ++j) {
		printf("%1.4f\n", res[j]);
	}
}