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.io.*; import java.util.*;
public class CF0449D {
   
    static int MXBIT=20;
    static int MAXN=1<<20;
    static int MOD=1000000007;
    static class FastReader 
    { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastReader() 
        { 
            br = new BufferedReader(new
                     InputStreamReader(System.in)); 
        } 
  
        String next() 
        { 
            while (st == null || !st.hasMoreElements()) 
            { 
                try
                { 
                    st = new StringTokenizer(br.readLine()); 
                } 
                catch (IOException  e) 
                { 
                    e.printStackTrace(); 
                } 
            } 
            return st.nextToken(); 
        } 
  
        int nextInt() 
        { 
            return Integer.parseInt(next()); 
        } 
  
        
    } 
    public static void main(String[] args) {
        int[] pow2=new int[MAXN+1];
        pow2[0]=1;
        for (int i = 1; i <=MAXN; i++) {
            pow2[i]=(pow2[i-1]*2)%MOD;
        }
        FastReader br=new FastReader();
        int N=br.nextInt();
        
        int[] dp=new int[MAXN];
        for (int i = 0; i < N; i++) {
            dp[br.nextInt()]++;
            //Evaluates f(mask,0)
        }
        for (int i = 0; i < MXBIT; i++) {
            for (int mask = 0; mask < MAXN; mask++) {//Memory compressed:
                //After processing i, everything is updated
                if((mask&(1<<i))==0){
                    dp[mask]+=dp[mask|(1<<i)];
                }
            }
        }
        int ans=0; 
        for (int mask = 0; mask < MAXN; mask++) {
            
            if(Integer.bitCount(mask)%2==1){
                ans-=pow2[dp[mask]]; 
            }else{
                ans+=pow2[dp[mask]]; 
            }
            ans%=MOD;
        }
        if(ans<0){
            ans+=MOD;
        }
        
        System.out.println(ans);
    }
    
}