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:
//package codeforces;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;

public class D {
    BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    PrintWriter writer = new PrintWriter(System.out);
    StringTokenizer stringTokenizer;

    String next() throws IOException {
        while (stringTokenizer == null || !stringTokenizer.hasMoreTokens()) {
            stringTokenizer = new StringTokenizer(reader.readLine());
        }
        return stringTokenizer.nextToken();
    }

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

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

    void solve() throws IOException {
        int n = nextInt();
        int[] f = new int[1 << 20];
        for(int i = 0; i < n; i++) {
            f[nextInt()]++;
        }
        for(int k = 0; k < 20; k++) {
            for(int i = 0; i < 1 << 20; i++) {
                int mask = 1 << k;
                if(((i & mask) > 0)) {
                    f[i ^ mask] += f[i];
                }
            }
        }
        int mod = 1000 * 1000 * 1000 + 7;
        int[] pow2 = new int[n + 1];
        pow2[0] = 1;
        for(int i = 1; i <= n; i++) {
            pow2[i] = pow2[i - 1] * 2;
            if(pow2[i] >= mod) {
                pow2[i] -= mod;
            }
        }
        long ans = 0;
        for(int i = 0; i < 1 << 20; i++) {
            if(Integer.bitCount(i) % 2 == 0) {
                ans += pow2[f[i]];
            } else {
                ans += mod - pow2[f[i]];
            }
        }
        writer.println(ans % mod);
        writer.close();
    }

    public static void main(String[] args) throws IOException {
        new D().solve();
    }
}