#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
#define pb push_back
#define all(a) a.begin(), a.end()
#define sz(a) ((int)a.size())
#ifdef Doludu
template <typename T>
ostream& operator << (ostream &o, vector <T> vec) {
    o << "{"; int f = 0;
    for (T i : vec) o << (f++ ? " " : "") << i;
    return o << "}";
}
void bug__(int c, auto ...a) {
    cerr << "\e[1;" << c << "m";
    (..., (cerr << a << " "));
    cerr << "\e[0m" << endl;
}
#define bug_(c, x...) bug__(c, __LINE__, "[" + string(#x) + "]", x)
#define bug(x...) bug_(32, x)
#define bugv(x...) bug_(36, vector(x))
#define safe bug_(33, "safe")
#else
#define bug(x...) void(0)
#define bugv(x...) void(0)
#define safe void(0)
#endif
const int mod = 998244353, N = 1 << 21;

int add(int a, int b) { a += b; if (a >= mod) a -= mod; return a; }
int sub(int a, int b) { a -= b; if (a < 0) a += mod; return a; }
int mul(int a, int b) { return 1ll * a * b % mod; }
int Pow(int a, int b) { int ans = 1; for (; b; b >>= 1, a = mul(a, a)) if (b & 1) ans = mul(ans, a); return ans; }

int main() {
    ios::sync_with_stdio(false), cin.tie(0);
    // 7 8 9 4
    // 4 5 6 1
    // 1 2 3 7

    // 1 2 3 4 5 6 7 10
    int t;
    cin >> t;
    while (t--) {
        int n, k;
        cin >> n >> k;
        if (k == 1 || k == n) {
            cout << "No\n";
        } else {
            cout << "Yes\n";
            vector a(3, vector(n, 0));
            for (int i = 0; i + 1 < n; ++i) a[0][i] = 2 * (n - 1) + i;
            for (int i = 0; i + 1 < n; ++i) a[1][i] = 1 * (n - 1) + i;
            for (int i = 0; i + 1 < n; ++i) a[2][i] = 0 * (n - 1) + i;
            a[0][n - 1] = a[1][0];
            a[1][n - 1] = a[2][0];
            a[2][n - 1] = a[0][0];
            reverse(a[0].begin(), a[0].begin() + n - k);
            reverse(a[1].begin(), a[1].begin() + n - k);
            reverse(a[2].begin(), a[2].begin() + n - k + 1);

            for (int i = 0; i < 3; ++i) {
                for (int j = 0; j < n; ++j) {
                    cout << a[i][j] + 1 << " \n"[j + 1 == n];
                }
            }
        }
    }
}