// Original C++ solution (slower):
#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cmath>

#include <climits>

#include <iostream>

#include <vector>

#include <string>

#include <sstream>

#include <algorithm>

#include <utility>

#include <set>

#include <map>

#include <stack>

#include <queue>

#include <deque>

#include <functional>



using namespace std;



#define fst first

#define scd second

#define PB push_back

#define MP make_pair

#define all(a) a.begin(),a.end()

#define rall(a) a.rbegin(),a.rend()

#define omajinai ios::sync_with_stdio(false);cin.tie(0)

#define rep(i,x) for(int i=0;i<(int)(x);++i)

#define rep1(i,x) for(int i=1;i<=(int)(x);++i)



using ll=long long;

using ld=long double;

using vi=vector<int>;

using vvi=vector<vi>;

using pii=pair<int, int>;

using vpii=vector<pii>;



template<class T,class U>ostream&operator<<(ostream&os,const pair<T,U>p){os<<"("<<p.fst<<", "<<p.scd<<")";return os;}

template<class T>ostream&operator<<(ostream&os,const vector<T>v){rep(i,v.size()){if(i)os<<", ";os<<v[i];}return os;}

template<typename T>T&max(T&a,T&b){if(a>=b)return a;return b;}

template<typename T>T&min(T&a,T&b){if(a < b)return a;return b;}

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

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



constexpr int inf = 1e9;

constexpr ll linf = 3e18;

constexpr double eps = 1e-9;

const int mod = 1e9 + 7;



ll dp[105][105][256];

ll fact[105];



int main()

{

    fact[0] = 1;

    for (int i = 1; i <= 100; ++i) {

        fact[i] = i * fact[i - 1];

        fact[i] %= mod;

    }



    int n, k; cin >> n >> k;



    vi a(n);



    rep(i, n) cin >> a[i];



    dp[0][0][0] = 1;



    for (int i = 0; i < n; ++i) {

        for (int j = 0; j <= i; ++j) {

            for (int k = 0; k <= 255; ++k) {

                (dp[i + 1][j + 1][k ^ a[i]] += dp[i][j][k]) %= mod;

                (dp[i + 1][j][k] += dp[i][j][k]) %= mod;

            }

        }

    }



    ll ans = 0;



    rep(i, n + 1) {

        ans += (ll)dp[n][i][k] * fact[i];

        ans %= mod;

    }



    cout << ans << endl;

}

// Optimized solution to the same problem (faster):
#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cmath>

#include <climits>

#include <iostream>

#include <vector>

#include <string>

#include <sstream>

#include <algorithm>

#include <utility>

#include <set>

#include <map>

#include <stack>

#include <queue>

#include <deque>

#include <functional>



using namespace std;



#define fst first

#define scd second

#define PB push_back

#define MP make_pair

#define all(a) a.begin(),a.end()

#define rall(a) a.rbegin(),a.rend()

#define omajinai ios::sync_with_stdio(false);cin.tie(0)

#define rep(i,x) for(int i=0;i<(int)(x);++i)

#define rep1(i,x) for(int i=1;i<=(int)(x);++i)



using ll=long long;

using ld=long double;

using vi=vector<int>;

using vvi=vector<vi>;

using pii=pair<int, int>;

using vpii=vector<pii>;



template<class T,class U>ostream&operator<<(ostream&os,const pair<T,U>p){os<<"("<<p.fst<<", "<<p.scd<<")";return os;}

template<class T>ostream&operator<<(ostream&os,const vector<T>v){rep(i,v.size()){if(i)os<<", ";os<<v[i];}return os;}

template<typename T>T&max(T&a,T&b){if(a>=b)return a;return b;}

template<typename T>T&min(T&a,T&b){if(a < b)return a;return b;}

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

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



constexpr int inf = 1e9;

constexpr ll linf = 3e18;

constexpr double eps = 1e-9;

const int mod = 1e9 + 7;



int dp[105][105][256];

ll fact[105];



int main()

{

    fact[0] = 1;

    for (int i = 1; i <= 100; ++i) {

        fact[i] = i * fact[i - 1];

        fact[i] %= mod;

    }



    int n, k; cin >> n >> k;



    vi a(n);



    rep(i, n) cin >> a[i];



    dp[0][0][0] = 1;



    for (int i = 0; i < n; ++i) {

        for (int j = 0; j <= i; ++j) {

            for (int k = 0; k <= 255; ++k) {

                (dp[i + 1][j + 1][k ^ a[i]] += dp[i][j][k]) %= mod;

                (dp[i + 1][j][k] += dp[i][j][k]) %= mod;

            }

        }

    }



    int ans = 0;



    rep(i, n + 1) {

        ans += (dp[n][i][k] * fact[i]) % mod;

        ans %= mod;

    }



    cout << ans << endl;

}


// Fastest known solution for this problem: