#include <iostream>
#include <vector>
#include <algorithm>
#include <map>
#include <string>
#include <queue>
#include <fstream>
#include <string>
#include <math.h>
#include <set>
#include <string>
#include <bitset>
#include <fstream>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> P;
//typedef pair<ll, P> T;

#define rep(i, n) for (ll i = 0; i < (n); i++)
#define revrep(i, n) for (ll i = (n) - 1; i >= 0; i--)
#define f first
#define s second
#define pb push_back

ll mod = 1000000007;
const ll INF = 100000000000010000;
//const ull B = 998244353;

//ll bit[200010];
//ll kai[3010];
//ll Cnt[10010];

//void pres(double A, ll x = 20) { printf("%.20f\n", x); }



/*ll sum(ll i)
{
	i += 1;
	ll t = 0;
	while (i > 0)
	{
		t += bit[i];
		i -= i & -i;
	}
	return t;
}

void add(ll i, ll x)
{
	i += 1;
	while (i <= 200005)
	{
		bit[i] += x;
		i += i & -i;
	}
}*/

/*`ll SAIKI(ll S) {
	for (ll T = S; ; T = (T - 1) & S) {

	}
}*/



void test() {
	cout << "888888888888888888888888" << endl;
}

ll gcd(ll x, ll y) {
	if (y == 0) return x;
	return gcd(y, x % y);
}

ll pow_mod(ll x, ll k) {
	x %= mod;
	ll res = 1;
	while (k > 0) {
		if (k % 2) {
			res *= x; res %= mod;
		}
		x *= x; x %= mod;
		k /= 2;
	}
	return res;
}

ll div_mod(ll x) {
	return pow_mod(x, mod - 2);
}


vector <ll> par; // 各元の親を表す配列
vector <ll> siz; // 素集合のサイズを表す配列(1 で初期化)

void init_u(ll sz_) {
	par.resize(sz_);
	siz.assign(sz_, 1LL);  // resize だとなぜか初期化されなかった
	for (ll i = 0; i < sz_; ++i) par[i] = i; // 初期では親は自分自身
}

// Member Function
// Find
ll root(ll x) { // 根の検索
	while (par[x] != x) {
		x = par[x] = par[par[x]]; // x の親の親を x の親とする
	}
	return x;
}

// Union(Unite, Merge)
bool merge(ll x, ll y) {
	x = root(x);
	y = root(y);
	if (x == y) return false;
	// merge technique（データ構造をマージするテク．小を大にくっつける）
	if (siz[x] < siz[y]) swap(x, y);
	siz[x] += siz[y];
	par[y] = x;
	return true;
}

bool issame(ll x, ll y) { // 連結判定
	return root(x) == root(y);
}

ll size(ll x) { // 素集合のサイズ
	return siz[root(x)];
}


//コンビネーション
const int MAXcomb = 1000010;
ll fac[MAXcomb], finv[MAXcomb], inv[MAXcomb];
//facはn!,finvは1/n!
//invは逆元
void COMinit() {
	fac[0] = fac[1] = 1;
	finv[0] = finv[1] = 1;
	inv[1] = 1;
	for (int i = 2; i < MAXcomb; i++) {
		fac[i] = fac[i - 1] * i % mod;
		inv[i] = mod - inv[mod%i] * (mod / i) % mod;
		finv[i] = finv[i - 1] * inv[i] % mod;
	}
}
ll comb(int n, int k) {
	if (n < k) return 0;
	if (n < 0 || k < 0) return 0;
	return fac[n] * finv[k] % mod * finv[n - k] % mod;
}


/*map<ll, ll> pres;
ll ct;

void makePrimeList(ll x) {
	ll X = x;
	for (ll j = 2; j * j <= x; j++) {
		ll cnt = 0;
		while (X % j == 0) {
			cnt++;
			X /= j;
		}
		if (cnt) {
			if (pres[j] == 0) {
				ct++;
				pres[j] = ct;
			}
			Cnt[pres[j]] += cnt;
		}
	}
	if (X != 1) {
		if (pres[X] == 0) {
			ct++;
			pres[X] = ct;
		}
		Cnt[pres[X]] += 1;
	}
}*/

ll n, k, a[101];
ll dp[101][256][101];
ll ans;
ll kai[101];

void solve() {
	cin >> n >> k;
	kai[0] = 1;
	rep(i, n) kai[i + 1] = kai[i] * (i + 1) % mod;
	rep(i, n) cin >> a[i];
	dp[0][0][0] = 1;
	rep(i, n) rep(j, 256) rep(k, n) {
		dp[i + 1][j][k] += dp[i][j][k];
		dp[i + 1][j][k] %= mod;
		dp[i + 1][j ^ a[i]][k + 1] += dp[i][j][k];
		dp[i + 1][j ^ a[i]][k + 1] %= mod;
	}
	rep(i, n + 1) {
		if (i == 0) continue;
		ll num = dp[n][k][i];
		ans += num * kai[i] % mod;
		ans %= mod;
	}
	cout << ans << endl;
}

int main() {
	
	cin.tie(0);
	ios::sync_with_stdio(false);
	ll T = 1;
	//cin >> T;
	rep(i, T) solve();
	ll N;
	cin >> N;
}
### Prompt

Create a solution in cpp for the following problem:

Sample testcase 3 has a mistake, so we erased this case and rejudged all solutions of this problem. (21:01)

Snuke got a sequence $a$ of length $n$ from AtCoder company. All elements in $a$ are distinct.
He made a sequence $b$, but actually, he is not remembered it.
However, he is remembered a few things about sequence $b$.


* All elements in $b$ are distinct.
* All elements in $b$ is in $a$.
* $b_1 \oplus b_2 \oplus \cdots \oplus b_r = k$. ($r$ is length of sequence $b$) [$\oplus$ means XOR]


For example, if $a = { 1, 2, 3 }$ and $k = 1$, he can make $b = { 1 }, { 2, 3 }, { 3, 2 }$.
He wants to restore sequence $b$, but he says that there are too many ways and he can't restore it. Please calculate the ways to make $b$ and help him.
Since the answer can be large, print the answer modulo $1,000,000,007$.

Input

The input is given from Standard Input in the following format:


> $n \ k$ $a_1 \ a_2 \ \cdots \ a_n$

Output

* Print the number of ways to make sequence $b$.
* Print the answer modulo $1,000,000,007$.



Constraints

* $1 \le n \le 100$
* $1 \le a_i, k \le 255$
* $i \neq j \Rightarrow a_i \neq a_j$



Subtasks

Subtask 1 [ $50$ points ]


* $1 \le n \le 4$

Subtask 2 [ $170$ points ]


* $1 \le n \le 20$

Subtask 3 [ $180$ points ]


* There are no additional constraints.

Output

* Print the number of ways to make sequence $b$.
* Print the answer modulo $1,000,000,007$.



Constraints

* $1 \le n \le 100$
* $1 \le a_i, k \le 255$
* $i \neq j \Rightarrow a_i \neq a_j$



Subtasks

Subtask 1 [ $50$ points ]


* $1 \le n \le 4$

Subtask 2 [ $170$ points ]


* $1 \le n \le 20$

Subtask 3 [ $180$ points ]


* There are no additional constraints.

Input

The input is given from Standard Input in the following format:


> $n \ k$ $a_1 \ a_2 \ \cdots \ a_n$

Examples

Input

3 1
1 2 3


Output

3


Input

3 10
8 7 5


Output

6


Input

25 127
5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 105 110 115 120 125


Output

235924722

### Response

```cpp
#include <iostream>
#include <vector>
#include <algorithm>
#include <map>
#include <string>
#include <queue>
#include <fstream>
#include <string>
#include <math.h>
#include <set>
#include <string>
#include <bitset>
#include <fstream>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> P;
//typedef pair<ll, P> T;

#define rep(i, n) for (ll i = 0; i < (n); i++)
#define revrep(i, n) for (ll i = (n) - 1; i >= 0; i--)
#define f first
#define s second
#define pb push_back

ll mod = 1000000007;
const ll INF = 100000000000010000;
//const ull B = 998244353;

//ll bit[200010];
//ll kai[3010];
//ll Cnt[10010];

//void pres(double A, ll x = 20) { printf("%.20f\n", x); }



/*ll sum(ll i)
{
	i += 1;
	ll t = 0;
	while (i > 0)
	{
		t += bit[i];
		i -= i & -i;
	}
	return t;
}

void add(ll i, ll x)
{
	i += 1;
	while (i <= 200005)
	{
		bit[i] += x;
		i += i & -i;
	}
}*/

/*`ll SAIKI(ll S) {
	for (ll T = S; ; T = (T - 1) & S) {

	}
}*/



void test() {
	cout << "888888888888888888888888" << endl;
}

ll gcd(ll x, ll y) {
	if (y == 0) return x;
	return gcd(y, x % y);
}

ll pow_mod(ll x, ll k) {
	x %= mod;
	ll res = 1;
	while (k > 0) {
		if (k % 2) {
			res *= x; res %= mod;
		}
		x *= x; x %= mod;
		k /= 2;
	}
	return res;
}

ll div_mod(ll x) {
	return pow_mod(x, mod - 2);
}


vector <ll> par; // 各元の親を表す配列
vector <ll> siz; // 素集合のサイズを表す配列(1 で初期化)

void init_u(ll sz_) {
	par.resize(sz_);
	siz.assign(sz_, 1LL);  // resize だとなぜか初期化されなかった
	for (ll i = 0; i < sz_; ++i) par[i] = i; // 初期では親は自分自身
}

// Member Function
// Find
ll root(ll x) { // 根の検索
	while (par[x] != x) {
		x = par[x] = par[par[x]]; // x の親の親を x の親とする
	}
	return x;
}

// Union(Unite, Merge)
bool merge(ll x, ll y) {
	x = root(x);
	y = root(y);
	if (x == y) return false;
	// merge technique（データ構造をマージするテク．小を大にくっつける）
	if (siz[x] < siz[y]) swap(x, y);
	siz[x] += siz[y];
	par[y] = x;
	return true;
}

bool issame(ll x, ll y) { // 連結判定
	return root(x) == root(y);
}

ll size(ll x) { // 素集合のサイズ
	return siz[root(x)];
}


//コンビネーション
const int MAXcomb = 1000010;
ll fac[MAXcomb], finv[MAXcomb], inv[MAXcomb];
//facはn!,finvは1/n!
//invは逆元
void COMinit() {
	fac[0] = fac[1] = 1;
	finv[0] = finv[1] = 1;
	inv[1] = 1;
	for (int i = 2; i < MAXcomb; i++) {
		fac[i] = fac[i - 1] * i % mod;
		inv[i] = mod - inv[mod%i] * (mod / i) % mod;
		finv[i] = finv[i - 1] * inv[i] % mod;
	}
}
ll comb(int n, int k) {
	if (n < k) return 0;
	if (n < 0 || k < 0) return 0;
	return fac[n] * finv[k] % mod * finv[n - k] % mod;
}


/*map<ll, ll> pres;
ll ct;

void makePrimeList(ll x) {
	ll X = x;
	for (ll j = 2; j * j <= x; j++) {
		ll cnt = 0;
		while (X % j == 0) {
			cnt++;
			X /= j;
		}
		if (cnt) {
			if (pres[j] == 0) {
				ct++;
				pres[j] = ct;
			}
			Cnt[pres[j]] += cnt;
		}
	}
	if (X != 1) {
		if (pres[X] == 0) {
			ct++;
			pres[X] = ct;
		}
		Cnt[pres[X]] += 1;
	}
}*/

ll n, k, a[101];
ll dp[101][256][101];
ll ans;
ll kai[101];

void solve() {
	cin >> n >> k;
	kai[0] = 1;
	rep(i, n) kai[i + 1] = kai[i] * (i + 1) % mod;
	rep(i, n) cin >> a[i];
	dp[0][0][0] = 1;
	rep(i, n) rep(j, 256) rep(k, n) {
		dp[i + 1][j][k] += dp[i][j][k];
		dp[i + 1][j][k] %= mod;
		dp[i + 1][j ^ a[i]][k + 1] += dp[i][j][k];
		dp[i + 1][j ^ a[i]][k + 1] %= mod;
	}
	rep(i, n + 1) {
		if (i == 0) continue;
		ll num = dp[n][k][i];
		ans += num * kai[i] % mod;
		ans %= mod;
	}
	cout << ans << endl;
}

int main() {
	
	cin.tie(0);
	ios::sync_with_stdio(false);
	ll T = 1;
	//cin >> T;
	rep(i, T) solve();
	ll N;
	cin >> N;
}
```