可以

#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1);
const double EPS = 1e-9;

typedef complex<double> cd;
const int MAXN = 1 << 19; // 2^19 = 524288 > 200000*20

cd A[MAXN], B[MAXN];
int rev[MAXN];

void init_fft(int n) {
    int bit = 0;
    while ((1 << bit) < n) bit++;
    for (int i = 0; i < n; i++) {
        rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (bit - 1));
    }
}

void fft(cd *a, int n, int inv) {
    for (int i = 0; i < n; i++) {
        if (i < rev[i]) swap(a[i], a[rev[i]]);
    }
    for (int len = 2; len <= n; len <<= 1) {
        double ang = 2 * PI / len * (inv ? -1 : 1);
        cd wn(cos(ang), sin(ang));
        for (int i = 0; i < n; i += len) {
            cd w(1);
            for (int j = 0; j < len / 2; j++) {
                cd u = a[i + j], v = a[i + j + len / 2] * w;
                a[i + j] = u + v;
                a[i + j + len / 2] = u - v;
                w *= wn;
            }
        }
    }
    if (inv) {
        for (int i = 0; i < n; i++) a[i] /= n;
    }
}

vector<double> poly_pow(vector<double> p, int y) {
    int n = 1;
    int max_sum = p.size() - 1;
    while (n <= max_sum * y) n <<= 1;
    init_fft(n);
    
    for (int i = 0; i < n; i++) A[i] = cd(0, 0);
    for (int i = 0; i <= max_sum; i++) A[i] = cd(p[i], 0);
    
    // 快速幂：log(y) 次 FFT
    for (int i = 0; i < n; i++) B[i] = cd(1, 0);
    int exp = y;
    while (exp > 0) {
        if (exp & 1) {
            fft(A, n, 0);
            fft(B, n, 0);
            for (int i = 0; i < n; i++) B[i] *= A[i];
            fft(B, n, 1);
            fft(A, n, 1);
            // 恢复 A
            for (int i = 0; i < n; i++) A[i] = cd(p[i % (max_sum+1)], 0);
        }
        fft(A, n, 0);
        for (int i = 0; i < n; i++) A[i] *= A[i];
        fft(A, n, 1);
        exp >>= 1;
    }
    
    vector<double> res(n);
    for (int i = 0; i < n; i++) res[i] = B[i].real();
    return res;
}

// 标准正态分布累积分布函数（使用误差函数，更精确）
double phi(double x) {
    return 0.5 * (1 + erf(x / sqrt(2)));
}

void solve() {
    int X, Y;
    cin >> X >> Y;
    
    // 大数据：中心极限定理
    if (Y > 800) {
        double mu = (X - 1) / 2.0;
        double var = (X * X - 1) / 12.0;
        double E = Y * mu;
        double sigma = sqrt(Y * var);
        
        for (int i = 0; i < 10; i++) {
            int A, B;
            cin >> A >> B;
            // 连续性校正
            double L = (A - 0.5 - E) / sigma;
            double R = (B + 0.5 - E) / sigma;
            double ans = phi(R) - phi(L);
            printf("%.6lf\n", ans);
        }
    } 
    // 小数据：FFT 精确计算
    else {
        // 单个骰子的概率分布
        vector<double> prob(X, 1.0 / X);
        
        // 计算 Y 次卷积
        vector<double> res = prob;
        for (int i = 2; i <= Y; i++) {
            vector<double> tmp(res.size() + X - 1, 0);
            for (int j = 0; j < res.size(); j++) {
                for (int k = 0; k < X; k++) {
                    tmp[j + k] += res[j] * prob[k];
                }
            }
            res = tmp;
        }
        
        // 前缀和以便快速查询区间概率
        vector<double> prefix(res.size() + 1, 0);
        for (int i = 0; i < res.size(); i++) {
            prefix[i+1] = prefix[i] + res[i];
        }
        
        for (int i = 0; i < 10; i++) {
            int A, B;
            cin >> A >> B;
            // 注意：伤害范围是 [A, B]，需要累加概率
            double ans = prefix[min(B, (int)res.size()-1) + 1] - prefix[A];
            printf("%.6lf\n", ans);
        }
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    
    int T;
    cin >> T;
    while (T--) {
        solve();
    }
    return 0;
}