2
1 31
 2
1 0
 10
450661 128600 993228 725823 293549 33490 843121 903634 556169 448234
 1
1
 5
1 3 5 7 9
 6
524 529 5249 524 529 529
 3
1 2 3
 2
0 1
 55
0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 2 3 4 5 6 7 8 9 3 4 5 6 7 8 9 4 5 6 7 8 9 5 6 7 8 9 6 7 8 9 7 8 9 8 9 9
 16
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
 3
128 1024 2048
 10
3 3 3 5 5 3 3 3 3 5
 1
0
 2
1 2
 2
0 0
 2
1 1

0
 2
 632
 0
 0
 24
 2
 2
 621247139
 64594
 4
 0
 1
 1
 3
 0

#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-10;
const double PI = atan(1.0) * 4;
const int inf = 2147483647 / 2;
const int maxn = 1000005;
const int mod = 1000000007;
template <class T>
void chmin(T &t, T first) {
  if (first < t) t = first;
}
template <class T>
void chmax(T &t, T first) {
  if (first > t) t = first;
}
template <class T>
int sgn(T first) {
  return (first > eps) - (first < -eps);
}
int an[maxn];
int f[1 << 20];
int two[1 << 20], siz[1 << 20];
int main() {
  two[0] = 1;
  siz[0] = 0;
  for (int i = (1); i <= (int)((1 << 20) - 1); i++)
    two[i] = two[i - 1] * 2 % mod;
  for (int i = (1); i <= (int)((1 << 20) - 1); i++)
    siz[i] = siz[i >> 1] ^ (i & 1);
  int n;
  while (cin >> n) {
    memset(f, 0, sizeof(f));
    for (int i = (0); i <= (int)(n - 1); i++) scanf("%d", &an[i]), ++f[an[i]];
    for (int i = (0); i <= (int)(20); i++)
      for (int j = ((1 << 20) - 1); j >= (int)(0); j--)
        if ((j >> i) & 1) f[j ^ (1 << i)] += f[j];
    int ans = 0;
    for (int i = (0); i <= (int)((1 << n) - 1); i++) {
      if (siz[i])
        ans += mod - two[f[i]] + 1;
      else
        ans += two[f[i]] - 1;
      if (ans >= mod) ans -= mod;
    }
    printf("%d\n", ans);
  }
  return 0;
}
