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 N = 1e6 + 5;
const int lim = 1e6;
const int mod = 1e9 + 7;
long long dp[(1 << 20) + 1], a[N];
long long binpow(long long u, long long p) {
  if (p == 0) return 1;
  if (p == 1) return u;
  long long t = binpow(u, p / 2);
  if (p & 1)
    return t * t % mod * u % mod;
  else
    return t * t % mod;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  if (fopen("A.inp", "r")) {
    freopen("test.inp", "r", stdin);
    freopen("test.ans", "w", stdout);
  }
  long long n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i], dp[a[i]]++;
  for (int i = 19; i >= 0; --i) {
    for (int j = 0; j <= lim; ++j) {
      if (j & (1 << i)) {
        dp[j ^ (1 << i)] += dp[j];
      }
    }
  }
  long long ans = 0;
  for (int i = 0; i <= lim; i++) {
    long long x = __builtin_popcount(i);
    if (x & 1) {
      ans = (ans - binpow(2ll, dp[i]) + 1 + mod) % mod;
    } else {
      ans = (ans + binpow(2ll, dp[i]) - 1 + mod) % mod;
    }
  }
  cout << ans;
}