def oi(): return int(input())
def os(): return input().rstrip()
def mi(): return list(map(int, input().split()))

# import sys
# input = sys.stdin.readline
# import sys
# sys.setrecursionlimit(10**8)
# import pypyjit
# pypyjit.set_param('max_unroll_recursion=-1')
input_count = 0
T = oi()
def solve(N,X,C):

    # 左右で各bitが埋まる最短の移動距離を調べる

    left = [0] * N
    right = [float("inf")] * N

    dictC = {}
    maxC = 0
    for i, c in enumerate(C):
        dictC[c] = i
        maxC = max(maxC, c)
    maxC += 1
    sums = 0
    for i, b in enumerate(bin(X)[2:].rjust(maxC, "0")[::-1]):
        if i in dictC:
            if b=="0":
                # これまでの和にどれだけ足せばいい？
                left[dictC[i]] = (pow(2,i) - sums)
            else:
                #今が1bit立っているということは0で達成できる
                left[dictC[i]] = 0
        sums += pow(2, i) * int(b)
        # print(sums)

    sums = 0
    for i, b in enumerate(bin(X)[2:].rjust(maxC, "0")[::-1]):
        if i in dictC:
            if b=="0":
                if X >= pow(2,i):
                    # これまでの和からどれだけ引けばいい？
                    right[dictC[i]] = sums+1
            else:
                right[dictC[i]] = 0
        sums += pow(2, i) * int(b)

    id_left = [(l,i) for i,l in enumerate(left)]
    out = float("inf")
    nexts = 0
    for a in sorted(id_left, reverse=True):
        if a[0] != 0:
            if out >= (nexts+a[0])*2:
                out = (nexts+a[0])*2
            else:
                break
        nexts = right[a[1]]

    if out == float("inf"):
        print(0)
    else:   
        print(out)
        # print(sums)
for _ in range(T):
    N, X = mi()
    C = mi()

    solve(N,X,C)