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;
int n, a;
long long int f[21][1000005], ans;
long long int fastexp(long long int b, long long int e) {
  long long int res = 1;
  while (e) {
    if (e & 1) res = (res * b) % 1000000007;
    e >>= 1;
    b = (b * b) % 1000000007;
  }
  return res;
}
long long int dp(int k, int x) {
  if (x > 1000000) return 0;
  if (f[k][x] == -1) {
    f[k][x] = dp(k - 1, x);
    if (!(x & (1 << (k - 1))))
      f[k][x] = (f[k][x] + dp(k - 1, x + (1 << (k - 1)))) % 1000000007;
  }
  return f[k][x];
}
int main() {
  scanf("%d", &n);
  memset(f, -1, sizeof f);
  memset(f[0], 0, sizeof f[0]);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a);
    f[0][a]++;
  }
  ans = 0;
  for (int i = 0; i < (1 << 20); i++)
    if (__builtin_popcount(i) & 1)
      ans = (ans - fastexp((long long int)2, dp(20, i)) + 1000000007) %
            1000000007;
    else
      ans = (ans + fastexp((long long int)2, dp(20, i))) % 1000000007;
  printf("%lld\n", ans);
  return 0;
}
