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

public class XorSeq {
	static final int MOD = (int) 1e9+ 7;
	public static int[][] mul(int[][] a, int[][] b){
		int[][] resM = new int[a.length][a.length];
		for (int i = 0; i < b.length; i++) {
			for (int j = 0; j < b.length; j++) {
				int res = 0;
				for (int k = 0; k < b.length; k++) {
//					System.out.printf("multiplying %d with %d\n", a[i][k], b[k][j]);
					res = (int) (res + (1L* a[i][k] * b[k][j]) % MOD) %MOD;
				}
//				System.out.println("res is "+res);
//				System.out.println("-------");
				resM[i][j] = res;
			}
		}
		return resM;
	}
	
	static int[][] matPow(int[][] base , long e){
		int[][] res = new int[base.length][base.length];
		for (int i = 0; i < res.length; i++) {
			res[i][i] = 1;
		}
		while(e>0){
			if((e&1) != 0)
				res = mul(res, base);
			base = mul(base,base);
			e>>=1;
		}
		return res;
	}
	
	public static void main(String[] args)throws IOException {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		long k = sc.nextLong();
		int[][] res = new int[n][n];
//		int[][] a = new int[n][n];
//		int[][] b = new int[n][n];
		long []a = new long[n];
		for (int i = 0; i < a.length; i++) {
			a[i] = sc.nextLong();
		}
		for (int i = 0; i < a.length; i++) {
			for (int j = i; j < a.length; j++) {
				if(Long.bitCount(a[i] ^ a[j])%3 == 0)
				{
					res[i][j] =1;
					res[j][i] = 1;
				}
			}
		}
		
		if(k>1){
			res = matPow(res, k-1);
			int ans = 0;
			for (int i = 0; i < a.length; i++) {
				for (int j = 0; j < a.length; j++) {
					ans= (ans + res[i][j]) % MOD;
				}
			}
			System.out.println(ans);
		}
		else
			System.out.println(n);
//		res = mul(a, b);
//		for (int i = 0; i < b.length; i++) {
//			for (int j = 0; j < b.length; j++) {
//				System.out.print(res[i][j]+" ");
//			}
//			System.out.println();
//		}
//		res = mul(res,)
	}
	
	
	
	
	
	
	static class Scanner 
	{
		StringTokenizer st;
		BufferedReader br;

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

		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();}


	}
}