Birthday Gift
Yarik's birthday is coming soon, and Mark decided to give him an array $$$a$$$ of length $$$n$$$.Mark knows that Yarik loves bitwise operations very much, and he also has a favorite number $$$x$$$, so Mark wants to find the maximum number $$$k$$$ such that it is possible to select pairs of numbers [$$$l_1, r_1$$$], [$$$l_2, r_2$$$], $$$\ldots$$$ [$$$l_k, r_k$$$], such that:   $$$l_1 = 1$$$.  $$$r_k = n$$$.  $$$l_i \le r_i$$$ for all $$$i$$$ from $$$1$$$ to $$$k$$$.  $$$r_i + 1 = l_{i + 1}$$$ for all $$$i$$$ from $$$1$$$ to $$$k - 1$$$.  $$$(a_{l_1} \oplus a_{l_1 + 1} \oplus \ldots \oplus a_{r_1}) | (a_{l_2} \oplus a_{l_2 + 1} \oplus \ldots \oplus a_{r_2}) | \ldots | (a_{l_k} \oplus a_{l_k + 1} \oplus \ldots \oplus a_{r_k}) \le x$$$, where $$$\oplus$$$ denotes the operation of bitwise XOR, and $$$|$$$ denotes the operation of bitwise OR. If such $$$k$$$ does not exist, then output $$$-1$$$.
In the first test case, you can take $$$k$$$ equal to $$$2$$$ and choose two segments [$$$1, 1$$$] and [$$$2, 3$$$], $$$(1) | (2 \oplus 3) = 1$$$. It can be shown that $$$2$$$ is the maximum possible answer.In the second test case, the segments [$$$1, 1$$$] and [$$$2, 2$$$] are suitable, $$$(1) | (1) = 1$$$. It is not possible to make more segments.In the third test case, it is not possible to choose $$$2$$$ segments, as $$$(1) | (3) = 3 > 2$$$, so the optimal answer is $$$1$$$.
Each test consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. The following lines contain the descriptions of the test cases.The first line of each test case contains two integers $$$n$$$ and $$$x$$$ ($$$1 \le n \le 10^5, 0 \le x < 2^{30}$$$) — the length of the array $$$a$$$ and the number $$$x$$$ respectively.The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \le a_i < 2^{30}$$$) — the array $$$a$$$ itself.It is guaranteed that the sum of the values of $$$n$$$ across all test cases does not exceed $$$10^5$$$.
For each test case, output a single integer on a separate line — the maximum suitable number $$$k$$$, and $$$-1$$$ if such $$$k$$$ does not exist.
def solve():
    t = int(input())
    for _ in range(t):
        (n, x) = map(int, input().split())
        x += 1
        a = list(map(int, input().split()))
        res = -1
        for i in range(30, -1, -1):
            b = []
            open = False
            for j in range(len(a)):
                if not open:
                    b.append(a[j])
                else:
                    b[-1] ^= a[j]
                if a[j] & 1 << i:
                    open = not open
            if not x & 1 << i:
                if open:
                    print(res)
                    break
                a = b
            elif not open:
                res = max(res, len(b))
        else:
            print(res)
solve()
def solve():
    (n, x) = map(int, input().split())
    x += 1
    a = list(map(int, input().split()))
    res = -1
    for i in range(30, -1, -1):
        b = []
        open = False
        for j in range(len(a)):
            if not open:
                b.append(a[j])
            else:
                b[-1] ^= a[j]
            if a[j] & 1 << i:
                open = not open
        if not x & 1 << i:
            if open:
                print(res)
                return
            a = b
        elif not open:
            res = max(res, len(b))
    print(res)
t = int(input())
for _ in range(t):
    solve()
def foo(n, x, a):
    val = 0
    ct = 0
    for z in a:
        z = int(z)
        val ^= z
        if val | x == x:
            ct += 1
            val = 0
    if val == 0:
        return ct
    else:
        return -1
t = int(input())
for _ in range(t):
    (n, x) = [int(z) for z in input().split()]
    a = [int(z) for z in input().split()]
    best = -1
    highbit = 1
    while highbit:
        best = max(best, foo(n, x + highbit - 1, a))
        highbit = x & -x
        x ^= highbit
    print(best)