Problem: You are given n integers a1, a2, ..., an.

A sequence of integers x1, x2, ..., xk is called a "xor-sequence" if for every 1 ≤ i ≤ k - 1 the number of ones in the binary representation of the number xi <image> xi + 1's is a multiple of 3 and <image> for all 1 ≤ i ≤ k. The symbol <image> is used for the binary exclusive or operation.

How many "xor-sequences" of length k exist? Output the answer modulo 109 + 7.

Note if a = [1, 1] and k = 1 then the answer is 2, because you should consider the ones from a as different.

Input

The first line contains two integers n and k (1 ≤ n ≤ 100, 1 ≤ k ≤ 1018) — the number of given integers and the length of the "xor-sequences".

The second line contains n integers ai (0 ≤ ai ≤ 1018).

Output

Print the only integer c — the number of "xor-sequences" of length k modulo 109 + 7.

Examples

Input

5 2
15 1 2 4 8


Output

13


Input

5 1
15 1 2 4 8


Output

5

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int M = 1e9 + 7;
struct Mint {
  int x;
  Mint(int _value = 0) : x(_value) {}
  Mint MMI() const {
    int ax = x, bx = M, sx = 1, tx = 0, qx;
    while (bx) {
      qx = ax / bx;
      ax -= qx * bx;
      sx -= qx * tx;
      swap(ax, bx);
      swap(sx, tx);
    }
    return sx >= 0 ? sx : sx + M;
  }
  operator int() { return x; }
  Mint operator-(const Mint& _a) const {
    int ans = x - _a.x;
    return ans < 0 ? M + ans : ans;
  }
  Mint operator+(const Mint& _a) const {
    int ans = x + _a.x;
    return ans - (ans >= M) * M;
  }
  Mint operator*(const Mint& _a) const { return (long long)x * _a.x % M; }
  Mint operator/(const Mint& _a) const { return (long long)x * _a.MMI().x % M; }
  bool operator==(const Mint& _a) const { return x == _a.x; }
  bool operator!=(const Mint& _a) const { return x != _a.x; }
};
Mint& operator+=(Mint& _a, const Mint& _b) { return _a = _a + _b; }
Mint& operator-=(Mint& _a, const Mint& _b) { return _a = _a - _b; }
Mint& operator*=(Mint& _a, const Mint& _b) { return _a = _a * _b; }
Mint& operator/=(Mint& _a, const Mint& _b) { return _a = _a / _b; }
istream& operator>>(istream& is, Mint& _a) {
  is >> _a.x;
  return is;
}
ostream& operator<<(ostream& os, Mint& _a) {
  os << _a.x;
  return os;
}
namespace Matrix {
template <typename T>
vector<vector<T>> Zero(int _n, int _m) {
  return vector<vector<T>>(_n, vector<T>(_m, 0));
}
template <typename T>
vector<vector<T>> Identity(int _n) {
  vector<vector<T>> ans = Zero<T>(_n, _n);
  for (int i = 0; i < _n; i++) ans[i][i] = 1;
  return ans;
}
template <typename T>
vector<vector<T>> Multiplication(const vector<vector<T>>& _a,
                                 const vector<vector<T>>& _b) {
  assert(_a[0].size() == _b.size());
  vector<vector<T>> ans = Zero<T>(_a.size(), _b[0].size());
  for (int i = 0; i < _a.size(); i++)
    for (int j = 0; j < _b[0].size(); j++)
      for (int k = 0; k < _a[0].size(); k++) ans[i][j] += _a[i][k] * _b[k][j];
  return ans;
}
template <typename T>
vector<vector<T>> Power(vector<vector<T>> _a, long long expo) {
  vector<vector<T>> ans = Identity<T>(_a.size());
  while (expo > 0) {
    if (expo & 1) ans = Multiplication(ans, _a);
    expo >>= 1;
    _a = Multiplication(_a, _a);
  }
  return ans;
}
}  // namespace Matrix
long long n, k;
vector<long long> a;
bool Read() {
  cin >> n;
  if (cin.eof()) return 0;
  cin >> k;
  a.resize(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  return 1;
}
void Process() {
  vector<vector<Mint>> exor(n, vector<Mint>(n));
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      exor[i][j] = __builtin_popcountll(a[i] ^ a[j]) % 3 == 0;
  const vector<vector<Mint>> p = Matrix::Power(exor, k - 1);
  Mint ans = 0;
  for (const auto& row : p)
    for (Mint x : row) ans += x;
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  while (Read()) Process();
  return 0;
}