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.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;

public class XorSequences {

	static final int MOD = (int) 1e9 + 7;

	static int[][] multMatrix(int[][] a, int[][] b) {
		int[][] res = new int[a.length][a.length];
		for (int i = 0; i < a.length; i++)
			for (int j = 0; j < a.length; j++) {
				int ans = 0;
				for (int k = 0; k < a.length; k++) {
					ans = (int) ((ans + (1L * a[i][k] * b[k][j]) % MOD) % MOD);
				}
				res[i][j] = ans;
			}
		return res;
	}

	static int[][] matrixExpo(int[][] base, long e) {
		int n = base.length;
		int[][] ans = new int[n][n];
		for (int i = 0; i < n; i++)
			ans[i][i] = 1;
		while (e != 0) {
			if ((e & 1) != 0)
				ans = multMatrix(ans, base);
			base = multMatrix(base, base);
			e >>= 1;
		}
		return ans;
	}

	static long sum(int[][] a) {
		long sum = 0;
		int n = a.length;
		for (int i = 0; i < n; i++)
			for (int j = 0; j < n; j++) {
				sum += a[i][j];
				if (sum > MOD)
					sum -= MOD;
			}
		return sum;
	}

	public static void main(String[] args) throws IOException {
		MyScanner sc = new MyScanner(System.in);
		PrintWriter pw = new PrintWriter(System.out);
		int n = sc.nextInt();
		long k = sc.nextLong();
		long[] a = new long[n];
		for (int i = 0; i < n; i++) {
			a[i] = sc.nextLong();
		}
		int[][] adjMatrix = new int[n][n];
		for (int i = 0; i < n; i++) {
			for (int j = i; j < n; j++) {
				if (Long.bitCount(a[i] ^ a[j]) % 3 == 0)
					adjMatrix[i][j] = adjMatrix[j][i] = 1;
			}
		}
		if (k > 1) {
			adjMatrix = matrixExpo(adjMatrix, k - 1);
			pw.println(sum(adjMatrix));
		} else
			pw.println(n);

		pw.close();
	}

	static class MyScanner {
		StringTokenizer st;
		BufferedReader br;

		public MyScanner(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();
		}

	}
}