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 JAVA):

Solution:
import java.util.*;
import java.io.*;

public class Main {
	public static void main(String args[]) {new Main().run();}

	FastReader in = new FastReader();
	PrintWriter out = new PrintWriter(System.out);
	void run(){
		work();
		out.flush();
	}
	long mod=1000000007L;
	long inf=Long.MAX_VALUE;
	long gcd(long a,long b) {
		return a==0?b:gcd(b%a,a);
	}
	void work() {
		int n=ni();
		long k=nl();
		long[] A=na(n);
		long[][] B=new long[1][n];
		Arrays.fill(B[0], 1);
		long[][] M=new long[n][n];
		for(int i=0;i<n;i++) {
			for(int j=0;j<n;j++) {
				int c=0;
				long v=A[i]^A[j];
				for(;v>0;v>>=1) {
					if(v%2==1) {
						c++;
					}
				}
				M[i][j]=c%3==0?1:0;
			}
		}
		k--;
		while(k>0) {
			if(k%2==1) {
				B=mul(B,M);
			}
			M=mul(M,M);
			k/=2;
		}
		long ret=0;
		for(long a:B[0]) {
			ret+=a;
			ret%=mod;
		}
		out.println(ret);
	}
	
	
	private long[][] mul(long[][] A, long[][] B) {
		int n=A.length,m=A[0].length,l=B[0].length;
		long[][] D=new long[n][l];
		for(int i=0;i<n;i++) {
			for(int j=0;j<l;j++) {
				for(int k=0;k<m;k++) {
					D[i][j]+=A[i][k]*B[k][j];
					D[i][j]%=mod;
				}
			}
		}
		return D;
	}
	//input
	@SuppressWarnings("unused")
	private ArrayList<Integer>[] ng(int n, int m) {
		ArrayList<Integer>[] graph=(ArrayList<Integer>[])new ArrayList[n];
		for(int i=0;i<n;i++) {
			graph[i]=new ArrayList<>();
		}
		for(int i=1;i<=m;i++) {
			int s=in.nextInt()-1,e=in.nextInt()-1;
			graph[s].add(e);
			graph[e].add(s);
		}
		return graph;
	}

	private ArrayList<long[]>[] ngw(int n, int m) {
		ArrayList<long[]>[] graph=(ArrayList<long[]>[])new ArrayList[n];
		for(int i=0;i<n;i++) {
			graph[i]=new ArrayList<>();
		}
		for(int i=1;i<=m;i++) {
			long s=in.nextLong()-1,e=in.nextLong()-1,w=in.nextLong();
			graph[(int)s].add(new long[] {e,w,i});
			graph[(int)e].add(new long[] {s,w});
		}
		return graph;
	}

	private int ni() {
		return in.nextInt();
	}

	private long nl() {
		return in.nextLong();
	}
	
	private String ns() {
		return in.next();
	}

	private long[] na(int n) {
		long[] A=new long[n];
		for(int i=0;i<n;i++) {
			A[i]=in.nextLong();
		}
		return A;
	}
	private int[] nia(int n) {
		int[] A=new int[n];
		for(int i=0;i<n;i++) {
			A[i]=in.nextInt();
		}
		return A;
	}
}	

class FastReader
{
	BufferedReader br;
	StringTokenizer st;

	public FastReader()
	{
		br=new BufferedReader(new InputStreamReader(System.in));
	}


	public String next() 
	{
		while(st==null || !st.hasMoreElements())//回车，空行情况
		{
			try {
				st = new StringTokenizer(br.readLine());
			} catch (IOException e) {
				e.printStackTrace();
			}
		}
		return st.nextToken();
	}

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

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