#include <bits/stdc++.h>
#define rep(i, p, n) for (ll i = p; i < (ll)(n); i++)
#define rep2(i, p, n) for (ll i = p; i >= (ll)(n); i--)
using namespace std;
using ll = long long;
using ld = long double;
const double pi = 3.141592653589793;
const long long inf = 2 * 1e9;
const long long linf = 4 * 1e18;
const ll mod1 = 1000000007;
const ll mod2 = 998244353;
template <class T>
inline bool chmax(T &a, T b)
{    if (a < b)    {        a = b;        return 1;    }    return 0;}
template <class T>
inline bool chmin(T &a, T b)
{
    if (a > b)
    {
        a = b;
        return 1;
    }
    return 0;
}

//atcoder
#include <atcoder/all>
using namespace atcoder;
using mint1 = modint1000000007;
using mint2 = modint998244353;

vector<pair<ll, ll>> base={{-1, 0}, {1, 0}, {0, -1}, {0, 1}};

int main() {

    //////////////////
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    //////////////////

    ll T;
    cin >> T;
    vector<ll> li(62);
    li.at(0) = 1;
    rep(i, 0, 61) {
        li.at(i + 1) = li.at(i) * 2;
    }
    while (T--)
    {
        ll N, X;
        cin >> N >> X;
        vector<ll> C(N);
        rep(i, 0, N) {
            cin >> C.at(i);
        }
        ll ans = 0;
        rep2(i, N - 1, 0)
        {
            if (li.at(C.at(i))&X) {
                continue;
            }
            //cout << C.at(i) << " " << i << endl;
            ans=linf;
            if (C.at(i)==0) {
                ans = 1;
                break;
            }
            ll temp=1;
	        if ((li.at(C.at(i)-1)&X)==0) {
        		rep(j, 0, C.at(i)) {
                	if (X&li.at(j)) {
                    	temp += li.at(j);
                	}
                }
                //cout << temp;
                if (X>=temp) {
                	chmin(ans, temp);	
                }
                temp=1;
            }
            rep(j, 0, C.at(i)) {
                if ((X&li.at(j))==0) {
                    temp += li.at(j);
                }
            }
            chmin(ans, temp);	
            break;
        }
        cout << ans*2 << endl;
    }
}