There is an array with n elements a1, a2, ..., an and the number x.

In one operation you can select some i (1 ≤ i ≤ n) and replace element ai with ai & x, where & denotes the [bitwise and](https://en.wikipedia.org/wiki/Bitwise_operation#AND) operation.

You want the array to have at least two equal elements after applying some operations (possibly, none). In other words, there should be at least two distinct indices i ≠ j such that ai = aj. Determine whether it is possible to achieve and, if possible, the minimal number of operations to apply.

Input

The first line contains integers n and x (2 ≤ n ≤ 100 000, 1 ≤ x ≤ 100 000), number of elements in the array and the number to and with.

The second line contains n integers ai (1 ≤ ai ≤ 100 000), the elements of the array.

Output

Print a single integer denoting the minimal number of operations to do, or -1, if it is impossible.

Examples

Input

4 3
1 2 3 7


Output

1


Input

2 228
1 1


Output

0


Input

3 7
1 2 3


Output

-1

Note

In the first example one can apply the operation to the last element of the array. That replaces 7 with 3, so we achieve the goal in one move.

In the second example the array already has two equal elements.

In the third example applying the operation won't change the array at all, so it is impossible to make some pair of elements equal.
#include <bits/stdc++.h>
using namespace std;
int xx[4] = {1, -1, 0, 0};
int yy[4] = {0, 0, 1, -1};
const double eps = 1e-9;
const int maxn = 2e6 + 5000;
const long long mod = 1e9 + 7;
const int INF(0x3f3f3f3f);
const long long INFL(0x3f3f3f3f3f3f3f3fll);
inline int sign(double a) { return a < -eps ? -1 : a > eps; }
long long mul(long long a, long long b, long long c) {
  long long res = 1;
  while (b) {
    if (b & 1) res *= a, res %= c;
    a *= a, a %= c, b >>= 1;
  }
  return res;
}
long long phi(long long x) {
  long long res = x;
  for (long long i = 2; i * i <= x; i++) {
    if (x % i == 0) res = res / i * (i - 1);
    while (x % i == 0) x /= i;
  }
  if (x > 1) res = res / x * (x - 1);
  return res;
}
template <typename A, typename B>
inline bool chmin(A& a, B b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <typename A, typename B>
inline bool chmax(A& a, B b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <typename A, typename B>
inline long long add(A x, B y) {
  if (x + y < 0) return x + y + mod;
  return x + y >= mod ? x + y - mod : x + y;
}
template <typename A, typename B>
inline void add2(A& x, B y) {
  if (x + y < 0)
    x = x + y + mod;
  else
    x = (x + y >= mod ? x + y - mod : x + y);
}
template <typename A, typename B>
inline long long mul1(A x, B y) {
  return 1ll * x * y % mod;
}
template <typename A, typename B>
inline void mul2(A& x, B y) {
  x = (1ll * x * y % mod + mod) % mod;
}
template <typename A>
inline void debug(A a) {
  cout << a << '\n';
}
template <typename A>
inline long long sqr(A x) {
  return 1ll * x * x;
}
template <typename A>
A inv(A x) {
  return mul(x, mod - 2);
}
template <class T>
void wt(const T& x) {
  cout << x << endl;
}
template <class T>
void wt(const T& x, char c) {
  cout << x << c;
}
template <class T>
void wt(const T& x, const string& s) {
  cout << x << s;
}
template <class T>
void wt(const T& x, int rnd) {
  cout << fixed << setprecision(rnd) << x << endl;
}
inline void debug() { cout << "###!!!" << endl; }
inline long long read() {
  char c = getchar();
  long long x = 0, f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = 1ll * x * 10 + c - '0', c = getchar();
  return x * f;
}
int fa[maxn];
int Find(int x) {
  if (x != fa[x]) return fa[x] = Find(fa[x]);
  return fa[x];
}
long long n, k;
int a[maxn];
int b[maxn];
string s[maxn];
int cnt[maxn];
map<int, int> m;
int main() {
  ios::sync_with_stdio(false);
  int x;
  while (cin >> n >> x) {
    for (int i = 1; i <= n; i++) cin >> a[i], m[a[i]]++;
    for (auto d : m) {
      if (d.second >= 2) {
        cout << 0 << endl;
        return 0;
      }
    }
    m.clear();
    m[x]++;
    string str = "";
    int flag = 0;
    for (int i = 1; i <= n; i++)
      if (m[a[i]]) flag = 1;
    map<int, int> mp;
    if (flag) {
      for (int i = 1; i <= n; i++) {
        if (m[a[i]]) continue;
        mp[a[i] & x]++;
        if (x == a[i] & x && mp[a[i] & x]) {
          cout << 1 << endl;
          return 0;
        }
      }
    }
    mp.clear();
    map<pair<int, int>, int> mp1;
    m.clear();
    for (int i = 1; i <= n; i++) mp1[pair<int, int>(a[i], i)]++;
    for (int i = 1; i <= n; i++) m[a[i]]++;
    for (int i = 1; i <= n; i++) {
      mp[a[i] & x]++;
      if (m[a[i] & x] && mp1[pair<int, int>(a[i] & x, i)] == 0) {
        cout << 1 << endl;
        return 0;
      }
    }
    for (auto d : mp) {
      if (d.second >= 2) {
        cout << 2 << endl;
        return 0;
      }
    }
    cout << -1 << endl;
  }
}
