Problem: Jzzhu have n non-negative integers a1, a2, ..., an. We will call a sequence of indexes i1, i2, ..., ik (1 ≤ i1 < i2 < ... < ik ≤ n) a group of size k. 

Jzzhu wonders, how many groups exists such that ai1 & ai2 & ... & aik = 0 (1 ≤ k ≤ n)? Help him and print this number modulo 1000000007 (109 + 7). Operation x & y denotes bitwise AND operation of two numbers.

Input

The first line contains a single integer n (1 ≤ n ≤ 106). The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 106).

Output

Output a single integer representing the number of required groups modulo 1000000007 (109 + 7).

Examples

Input

3
2 3 3


Output

0


Input

4
0 1 2 3


Output

10


Input

6
5 2 0 5 2 1


Output

53

Solution (in JAVA):

Solution:

import java.util.Scanner;


public class Main {
    private static final int mod = (int)1e9+7;
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        try{

        Scanner sc = new Scanner(System.in);
        String input1 = sc.nextLine();
        String input2 = sc.nextLine();
        
        int n = Integer.parseInt(input1);
        
        String str1[] = input2.split(" ");
        int xs[] = new int[n];
        
        for(int i=0; i<str1.length; i++){
            xs[i] = Integer.parseInt(str1[i].trim());
        }
        
        int[] pow2 = new int[n+1];
        pow2[0] = 1;
        for(int i = 1; i < pow2.length; i++) pow2[i] = (int)(pow2[i-1] * 2L % mod);
        
        int[] dp = new int[1<<20];
        for(int x : xs) {
            dp[x]++;
        }
        
        
        for(int i = 0; i < 20; i++) {
            for(int j = 0; j < 1 << 20; j++) {
                if((j>>>i&1) == 0) {
                    dp[j] += dp[j|1<<i];
                }
            }
        }
        
        long ans = 0;
        for(int i = 0; i < 1 << 20; i++) {
            ans += (1L-Integer.bitCount(i)%2*2) * pow2[dp[i]]%mod;
        }
        
        System.out.println((ans % mod + mod) % mod);
        
        }
        catch(Exception e){
            e.printStackTrace();
        }
    }

}