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 B = 1e9 + 7;
const int N = 1 << 20;
int n;
int f[21][N];
long long mypow(long long a, int b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a % B;
    a = a * a % B;
    b >>= 1;
  }
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = 0, a; i < n; ++i) {
    scanf("%d", &a);
    if (a & 1) {
      f[0][a]++;
      f[0][a ^ 1]++;
    } else {
      f[0][a]++;
    }
  }
  for (int i = 1; i <= 20; ++i) {
    for (int x = 0; x < N; ++x) {
      if (x >> i & 1) {
        f[i][x] = f[i - 1][x];
      } else {
        f[i][x] = f[i - 1][x] + f[i - 1][x + (1 << i)];
      }
    }
  }
  long long ans = 0;
  for (int x = 0; x < N; ++x) {
    if (__builtin_popcount(x) & 1) {
      ans = (ans + B - mypow(2, f[20][x])) % B;
    } else {
      ans = (ans + mypow(2, f[20][x])) % B;
    }
  }
  printf("%lld\n", ans);
  return 0;
}
