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
#include <bits/stdc++.h>
using namespace std;
const bool read_file = false;
const char fi[] = "XX.INP";
const char fo[] = "XX.OUT";
void set_up() {
  if (read_file) {
    freopen(fi, "r", stdin);
    freopen(fo, "w", stdin);
  }
  cin.clear();
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
void just_do_it();
int main() {
  set_up();
  just_do_it();
}
const int ms = 1e2;
const int mod = 1e9 + 7;
struct matrix {
  int rows;
  int cols;
  long long m[ms][ms];
  matrix() { memset(m, 0, sizeof(m)); }
};
void multiply(matrix &a, matrix &b) {
  int m = a.rows;
  int n = a.cols;
  int p = b.cols;
  matrix c;
  c.rows = m;
  c.cols = p;
  for (int i = 0; i < m; i++) {
    for (int j = 0; j < p; j++) {
      c.m[i][j] = 0;
      for (int k = 0; k < n; k++) {
        c.m[i][j] +=
            (long long)(a.m[i][k] % mod) * (long long)(b.m[k][j] % mod) % mod;
        c.m[i][j] %= mod;
      }
    }
  }
  a = c;
}
matrix t;
matrix h;
matrix res;
matrix tmp;
long long bits(long long x) {
  long long res = 0;
  while (x != 0) {
    res += x % 2;
    x /= 2;
  }
  return res;
}
bool check(long long x, long long y) { return bits(x ^ y) % 3 == 0; }
void power(long long x) {
  if (x == 1) {
    res = t;
    return;
  }
  power(x / 2);
  tmp = res;
  multiply(tmp, res);
  if (x % 2 == 1) {
    multiply(tmp, t);
  }
  res = tmp;
  return;
}
void just_do_it() {
  long long n, k;
  cin >> n >> k;
  t.rows = n;
  t.cols = n;
  h.rows = n;
  h.cols = 1;
  for (int i = 0; i < n; i++) {
    h.m[i][0] = 1;
  }
  long long a[n];
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  if (k == 1) {
    cout << n;
    return;
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      t.m[i][j] = check(a[i], a[j]);
    }
  }
  power(k - 1);
  multiply(res, h);
  long long s = 0;
  for (int i = 0; i < n; i++) {
    s += res.m[i][0] % mod;
    s %= mod;
  }
  cout << s;
}
