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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1000000007LL;
static int N;
static long long ans = 0, bin[1000007] = {1}, dp[1000007];
int main() {
  for (int i = 1; i < 1000007; i++) bin[i] = bin[i - 1] * 2 % MOD;
  scanf("%d", &N);
  for (int i = 0, x; i < N; i++) scanf("%d", &x), dp[x]++;
  for (int j = 0; j < 25; j++)
    for (int i = 0; i < 1000007; i++)
      if ((1 << j) & i) (dp[i ^ (1 << j)] += dp[i]) %= MOD;
  for (int i = 0; i < 1000007; i++) {
    int sign = 1;
    for (int j = 0; j < 25; j++)
      if ((1 << j) & i) sign = -sign;
    (ans += sign * (bin[dp[i]] - 1) % MOD + MOD) %= MOD;
  }
  printf("%lld\n", ans);
  return 0;
}