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 int inf = 1 << 28;
const int mod = (int)1e9 + 7;
int n, m, k;
int dp[1 << 20];
long long p[1000010];
int main() {
  cin >> n;
  int x;
  p[0] = 1;
  for (int i = 1; i <= n; i++) {
    p[i] = p[i - 1] * 2 % mod;
    scanf("%d", &x);
    dp[x]++;
  }
  for (int i = 0; i < 20; i++) {
    for (int j = 0; j < (1 << 20); j++) {
      if (j >> i & 1) dp[j - (1 << i)] += dp[j];
    }
  }
  long long ans = p[n];
  for (int i = 1; i < (1 << 20); i++) {
    int num = __builtin_popcount(i);
    if (num & 1)
      ans += mod - p[dp[i]];
    else
      ans += p[dp[i]];
    if (ans >= mod) ans -= mod;
  }
  cout << ans << endl;
  return 0;
}