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 long long mod = 1e9 + 7;
long long n, dp[20][1LL << 20], ans, po[(1LL << 20)];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  po[0] = 1;
  for (long long i = 1; i < (1LL << 20); ++i) po[i] = (po[i - 1] * 2) % mod;
  cin >> n;
  for (long long i = 1; i <= n; ++i) {
    long long a;
    cin >> a;
    for (long long j = 19; j >= 0; --j) {
      if (a & (1LL << j)) break;
      dp[j][0]++;
      dp[j][0] %= mod;
    }
    if (a % 2) {
      dp[0][a]++;
      dp[0][a] %= mod;
      dp[0][a ^ 1]++;
      dp[0][a ^ 1] %= mod;
    } else
      dp[0][a] = (dp[0][a] + 1) % mod;
  }
  for (long long i = 1; i < 20; ++i) {
    for (long long j = 1; j < (1LL << 20); ++j) {
      if ((j & (1LL << i)))
        dp[i][j] = dp[i - 1][j] % mod;
      else
        dp[i][j] = (dp[i - 1][j] + dp[i - 1][j + (1LL << i)]) % mod;
    }
  }
  for (long long i = 0; i < (1LL << 20); ++i) {
    long long rem = __builtin_popcount(i);
    rem %= mod;
    if (rem % 2 == 0) {
      ans += po[dp[19][i]];
      ans %= mod;
    } else {
      ans -= po[dp[19][i]];
      ans %= mod;
    }
  }
  cout << ans;
  return 0;
}
