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 maxn = (1 << 20) + 1e5;
const long long mode = 1e9 + 7;
int dp[maxn];
long long powerFun(int fi, int se) {
  if (se == 0) return 1;
  long long ret = powerFun(fi, se / 2);
  ret = (ret * ret) % mode;
  if (se % 2 == 0)
    return ret;
  else
    return (ret * 2) % mode;
}
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    dp[x]++;
  }
  for (int i = 0; i < 20; i++)
    for (int mask = (1 << 20) - 1; mask >= 0; mask--)
      if (mask & (1 << i)) dp[mask ^ (1 << i)] += dp[mask];
  long long ans = powerFun(2, n);
  for (int mask = 1; mask < (1 << 20); mask++) {
    if (__builtin_popcount(mask) % 2 == 1)
      ans -= powerFun(2, dp[mask]);
    else
      ans += powerFun(2, dp[mask]);
    if (ans >= mode) ans -= mode;
    if (ans < 0) ans += mode;
  }
  cout << ans << endl;
}