#include <bits/stdc++.h>
using namespace std;
#include <atcoder/all>
using namespace atcoder;
using mint = atcoder::modint998244353;
// using mint = double;

#define rep(i, l, r) for (int i = (int)(l); i<(int)(r); i++)
#define ll long long
#define ld long double
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define siz(x) (int)(x).size()

template<class T> bool chmin(T& a, T b) { if (a > b) {a = b; return true;} return false; }
template<class T> bool chmax(T& a, T b) { if (a < b) {a = b; return true;} return false; }

const int inf = 1e9;
const ll INF = 4e18;

template<class T> using pq = priority_queue<T, vector<T>, less<T>>;
template<class T> using spq = priority_queue<T, vector<T>, greater<T>>;

vector<int> di = {0, 0, 1, -1};
vector<int> dj = {1, -1, 0, 0};

struct Edge {
    int to, cost;
};

void solve() {
    int N, L; cin >> N >> L;
    vector<int> D(N);
    for (int& d : D) cin >> d;

    int cnt = 0;
    if (L%2 == 0) {
        rep(i, 0, N) {
            if (D[i] < L / 2) {
                if (binary_search(all(D), D[i] + L / 2)) cnt++;
            }
        }
    }
    // cout << cnt << endl;

    vector<mint> inv(N+1);
    rep(i, 1, N+1) {
        inv[i] = mint(1) / i;
    }

    if (cnt <= 1) {
        //NG
        mint ans = 0;
        //全部埋めるまでにかかる回数
        for (int i = 1; i <= N; i++) ans += N * inv[i];
        cout << ans.val() << endl;
        return;
    }

    vector dp(cnt + 1, vector<mint>(3));
    for (int i = cnt; i >= 0; i--) {
        for (int j = 2; j >= 0; j--) {
            if (j > i) continue;
            if (j == 2) {
                dp[i][j] = 0;
                continue;
            }
            int zero = cnt - i;
            int one = i - j;
            int two = j;

            mint blank = 2 * zero + one;

            //成功確率
            mint p = blank / N;
            //コスト
            mint cost = 1 / p;

            //0 -> 1
            if (zero > 0) dp[i][j] += dp[i+1][j] * mint(2 * zero) / blank;
            //1 -> 0
            if (one > 0) dp[i][j] += dp[i][j+1] * mint(one) / blank;
            dp[i][j] += cost;
        }
    }
    cout << dp[0][0].val() << endl;
}

int main() {
    int T; cin >> T;
    while(T--) solve();
}