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 int mod = 1e9 + 7;
int n, t, f[21][1 << 20];
int mus(int x, int y) { return (x - y + mod) % mod; }
int pow2(int n) {
  int t = 2, ans = 1;
  while (n) {
    if (n & 1) ans = (ans * t) % mod;
    t = (t * t) % mod;
    n >>= 1;
  }
  return ans;
}
void getf() {
  for (int i = 1; i <= 20; ++i) {
    for (int j = 0; j < (1 << 20); ++j) {
      if (j & (1 << (i - 1)))
        f[i][j] += f[i - 1][j];
      else
        f[i][j] += f[i - 1][j] + f[i - 1][j | (1 << (i - 1))];
    }
  }
}
int solve() {
  getf();
  int ans = pow2(n);
  for (int i = 1; i < (1 << 20); ++i) {
    int g = 0;
    for (int j = 0; j < 20; ++j)
      if (i & (1 << j)) g++;
    if (g & 1)
      ans = mus(ans, pow2(f[20][i]));
    else
      ans = (ans + pow2(f[20][i])) % mod;
  }
  return ans;
}
int main(void) {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &t);
    f[0][t]++;
  }
  printf("%d\n", solve());
}
