Subarray XOR less than K

/**
 *    author:  compounding
 *    created: 2024-11-13 01:43:23
 **/
#include <bits/stdc++.h>
using namespace std;
mt19937_64 RNG(chrono::steady_clock::now().time_since_epoch().count());
#define NeedForSpeed                  \
    ios_base::sync_with_stdio(false); \
    cin.tie(NULL);                    \
    cout.tie(NULL);
#define int long long
#define all(x) (x).begin(), (x).end()
typedef vector<int> vi;
typedef vector<bool> vb;
typedef vector<vi> vvi;
typedef vector<pair<int, int>> vpi;
#define f first
#define s second
#define yes cout << "YES" << endl
#define no cout << "NO" << endl
#define endl "\n"
const int mod = 1000000007;
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
struct trie
{
    int leftcnt, rightcnt; // count of left and right children which have 0, 1 in the bit
    trie *left, *right;    // reference to left and right child
    trie()
    {
        leftcnt = 0;
        rightcnt = 0;
        left = NULL;
        right = NULL;
    }
};
void insert(trie *root, int value)
{
    for (int i = 31; i >= 0; i--)
    {
        int biton = (value >> i) & 1; // get the ith bit
        // if ON
        if (biton)
        {
            root->rightcnt++;
            if (root->right == NULL)
            {
                // go to it's right child
                root->right = new trie();
            }
            root = root->right;
        }
        else
        {
            root->leftcnt++;
            if (root->left == NULL)
            {
                // go to it's left child
                root->left = new trie();
            }
            root = root->left;
        }
    }
}
int query(trie *root, int value, int k)
{
    int ans = 0;
    if (root == NULL)
    {
        return ans;
    }
    for (int i = 31; i >= 0; i--)
    {
        int biton = (value >> i) & 1; // get the ith bit
        int bitk = (k >> i) & 1;      // get the ith bit
        // TODO: if ith bit of k is ON
        if (bitk)
        {
            // TODO : if ith bit of value is ON
            if (biton)
            {
                // add the right cnt, as we have to make the XOR < k so we have to switch this bit off
                ans += root->rightcnt;
                if (root->left == NULL)
                {
                    return ans;
                }
                root = root->left;
            }
            else
            {
                ans += root->leftcnt;
                if (root->right == NULL)
                {
                    return ans;
                }
                root = root->right;
            }
        }
        // TODO: if ith bit of k is OFF
        else
        {
            // if ith bit of value is ON
            if (biton)
            {
                if (root->right == NULL)
                {
                    return ans;
                }
                root = root->right;
            }
            else
            {
                if (root->left == NULL)
                {
                    return ans;
                }
                root = root->left;
            }
        }
    }
    return ans;
}
void solve()
{
    int n, k;
    cin >> n >> k;
    // number of subarrays with XOR less than k
    vi a(n);
    int prexor = 0, cnt = 0;
    trie *root = new trie();
    insert(root, 0);
    for (int i = 0; i < n; i++)
    {
        cin >> a[i];
        prexor ^= a[i];
        cnt += query(root, prexor, k);
        insert(root, prexor);
    }
    cout << cnt << endl;
    return;
}

signed main()
{
    NeedForSpeed;
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}