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>
#pragma GCC optimize(2)
using namespace std;
int n, a[8800000], bin[8800000], ans[8800000], ss[8800000];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  bin[0] = 1;
  for (int i = 1; i <= n; ++i) bin[i] = bin[i - 1] * 2 % 1000000007;
  int sum = 0;
  for (int i = 0; i <= 20; ++i) {
    memset(ans, 0, sizeof ans);
    memset(ss, 0, sizeof ss);
    int ans1 = 0;
    for (int j = 1; j <= n; ++j) {
      if ((1 << i) & a[j]) {
        ans1++;
        ss[a[j]]++;
      }
    }
    for (int j = 0; j <= (1 << 21) - 1; ++j) {
      for (int k = 0; k <= 20; ++k)
        if (!(i & (1 << k))) ss[j] += ss[j | (1 << k)];
    }
    ss[0] = 0;
    int yu = bin[ans1] - 1;
    for (int j = 0; j <= (1 << (i + 1)) - 1; ++j) {
      if (j & (1 << i)) continue;
      int cnt = 1;
      for (int k = 0; k <= 20; ++k)
        if (j & (1 << k)) cnt *= -1;
      yu = (yu + cnt * (bin[ss[j]] - 1) + 1000000007) % 1000000007;
    }
    sum = (sum + yu) % 1000000007;
  }
  printf("%d\n", (bin[n] - 1 - sum + 1000000007) % 1000000007);
  return 0;
}
