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

 import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;


public class E {

	static final int mod = (int)1e9 + 7;
	public static void main(String[] args) throws IOException 
	{
		Scanner sc = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);

		int n = sc.nextInt();
		long a[] = new long[n];
		long k = sc.nextLong();
		for(int i = 0; i < n; ++i)
		a[i] = sc.nextLong();

		int[][] adjMat = new int[n][n];
		for(int i = 0; i < n; ++i)
			for(int j = 0; j < n; ++j)
				if(Long.bitCount(a[i] ^ a[j]) % 3 == 0)
					adjMat[i][j] = 1;
		int[][] M = matPow(adjMat, k - 1);

		int ans = 0;
		for(int i = 0; i < n; ++i)
			for(int j = 0; j < n; ++j)
			{
			    ans += M[i][j];
			    if(ans >= mod)
			        ans -= mod;
			}
		out.println(ans);
		out.flush();
		out.close();
	}
	

	static int[][] matPow(int[][] base, long p)
	{
		int n = base.length;
		int[][] ans = new int[n][n];
		for(int i = 0; i < n; i++)
			ans[i][i] = 1;
		while(p != 0)
		{
			if((p & 1) == 1)
				ans = matMul(ans, base, n, n, n);
			base = matMul(base, base, n, n, n);
			p >>= 1;
		}
		
		return ans;
	}
	
	static int[][] matMul(int[][] A, int[][] B, int p, int q, int r)	//C(p x r) = A(p x q) x (q x r) -- O(p x q x r)
	{
		int[][] C = new int[p][r];
		for(int i = 0; i < p; ++i)
			for(int j = 0; j < r; ++j)
				{
				    int x = 0;
				    for(int k = 0; k < q; ++k)
    					x = (int) ((x + 1l * A[i][k] * B[k][j] % mod) % mod);
    				C[i][j] = x;
				}
		return C;
	}

	static class Scanner 
	{
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s){    br = new BufferedReader(new InputStreamReader(s));}

		public Scanner(FileReader r){    br = new BufferedReader(r);}

		public String next() throws IOException 
		{
			while (st == null || !st.hasMoreTokens()) 
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public int nextInt() throws IOException {return Integer.parseInt(next());}

		public long nextLong() throws IOException {return Long.parseLong(next());}

		public String nextLine() throws IOException {return br.readLine();}

		public double nextDouble() throws IOException
		{
			String x = next();
			StringBuilder sb = new StringBuilder("0");
			double res = 0, f = 1;
			boolean dec = false, neg = false;
			int start = 0;
			if(x.charAt(0) == '-')
			{
				neg = true;
				start++;
			}
			for(int i = start; i < x.length(); i++)
				if(x.charAt(i) == '.')
				{
					res = Long.parseLong(sb.toString());
					sb = new StringBuilder("0");
					dec = true;
				}
				else
				{
					sb.append(x.charAt(i));
					if(dec)
						f *= 10;
				}
			res += Long.parseLong(sb.toString()) / f;
			return res * (neg?-1:1);
		}

		public boolean ready() throws IOException {return br.ready();}


	}
}