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 nmax = (1 << 20) + 5;
const int mod = 1e9 + 7;
int n, i, j, x, lim, sol;
int f[nmax], cnt[nmax], two[nmax];
void fix(int &x) {
  if (x < 0) x += mod;
  if (x >= mod) x -= mod;
}
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &x);
    f[x]++;
  }
  lim = 1 << 20;
  two[0] = 1;
  for (i = 1; i < lim; i++) two[i] = (two[i - 1] * 2) % mod;
  for (i = 0; i < 20; i++)
    for (j = 0; j < lim; j++) {
      if ((1 << i) & j)
        cnt[j]++;
      else
        f[j] += f[j + (1 << i)];
    }
  for (i = 1; i < lim; i++) {
    if (cnt[i] & 1) {
      sol += two[f[i]] - 1;
      fix(sol);
    } else {
      sol -= two[f[i]] - 1;
      fix(sol);
    }
  }
  sol = two[n] - 1 - sol;
  fix(sol);
  printf("%d\n", sol);
  return 0;
}
