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;
struct Matrix {
  static const int MAXN = 128;
  int r, c;
  long long a[MAXN][MAXN];
  long long* operator[](int i) { return a[i]; }
  const long long* operator[](int i) const { return a[i]; }
  void init(int row, int col) {
    this->r = row;
    this->c = col;
  }
  void fill(int row, int col, long long x = 0) {
    init(row, col);
    for (int i = 0; i < r; ++i) {
      ::fill(a[i], a[i] + c, x);
    }
  }
  void eye(int n, long long x = 1) {
    fill(n, n, 0);
    for (int i = 0; i < n; ++i) {
      a[i][i] = x;
    }
  }
  void set(const Matrix& o) {
    init(o.r, o.c);
    for (int i = 0; i < r; ++i) {
      copy(o[i], o[i] + c, a[i]);
    }
  }
};
void add(const Matrix& a, const Matrix& b, long long m, Matrix& ret) {
  static Matrix c;
  c.set(a);
  for (int i = 0; i < b.r; ++i) {
    for (int j = 0; j < b.c; ++j) {
      c[i][j] += b[i][j];
      if (c[i][j] >= m) {
        c[i][j] -= m;
      }
    }
  }
  ret.set(c);
}
void mul(const Matrix& a, const Matrix& b, long long m, Matrix& ret) {
  static Matrix c;
  c.init(a.r, b.c);
  for (int i = 0; i < c.r; ++i) {
    for (int j = 0; j < c.c; ++j) {
      long long x = 0;
      for (int k = 0; k < a.c; ++k) {
        x += a[i][k] * b[k][j] % m;
      }
      c[i][j] = x % m;
    }
  }
  ret.set(c);
}
void pow(const Matrix& a, long long b, long long m, Matrix& ret) {
  static Matrix c;
  c.set(a);
  ret.eye(c.r);
  while (b > 0) {
    if (b % 2 != 0) {
      mul(ret, c, m, ret);
    }
    b /= 2;
    mul(c, c, m, c);
  }
}
const int N = 111;
const int MOD = (int)(1e9 + 7);
int n;
long long k, a[N];
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; ++i) cin >> a[i];
  Matrix b;
  b.init(n, n);
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      if (__builtin_popcountll(a[i] ^ a[j]) % 3 == 0) {
        b[i][j] = 1;
      } else {
        b[i][j] = 0;
      }
    }
  }
  Matrix c;
  pow(b, k - 1, MOD, c);
  Matrix a;
  a.init(n, 1);
  for (int i = 0; i < n; ++i) a[i][0] = 1;
  Matrix d;
  mul(c, a, MOD, d);
  long long ans = 0;
  for (int i = 0; i < n; ++i) ans = (ans + d[i][0]) % MOD;
  cout << ans << endl;
  return 0;
}