Problem: Jzzhu have n non-negative integers a1, a2, ..., an. We will call a sequence of indexes i1, i2, ..., ik (1 ≤ i1 < i2 < ... < ik ≤ n) a group of size k. 

Jzzhu wonders, how many groups exists such that ai1 & ai2 & ... & aik = 0 (1 ≤ k ≤ n)? Help him and print this number modulo 1000000007 (109 + 7). Operation x & y denotes bitwise AND operation of two numbers.

Input

The first line contains a single integer n (1 ≤ n ≤ 106). The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 106).

Output

Output a single integer representing the number of required groups modulo 1000000007 (109 + 7).

Examples

Input

3
2 3 3


Output

0


Input

4
0 1 2 3


Output

10


Input

6
5 2 0 5 2 1


Output

53

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
ostream& operator<<(ostream& out, vector<T> const& a) {
  out << "[";
  for (int i = 0; i < int((a).size()); ++i) {
    if (i != 0) {
      out << ", ";
    }
    out << a[i];
  }
  out << "]";
  return out;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& out, pair<T1, T2> const& p) {
  out << "(" << p.first << ", " << p.second << ")";
  return out;
}
const int N = 1e6;
struct Data {
  int n;
  int a[N];
  bool read() {
    if (scanf("%d", &n) != 1) {
      return false;
    }
    for (int i = 0; i < n; ++i) {
      scanf("%d", &a[i]);
    }
    return true;
  }
  long long ans;
  void write() { printf("%d\n", (int)ans); }
  virtual void solve() {}
  virtual void clear() { *this = Data(); }
  void gen() {
    n = 1e6;
    for (int i = 0; i < n; ++i) {
      a[i] = i;
    }
  }
};
const int mod = 1e9 + 7;
int sum(int x, int y) {
  x += y;
  if (x >= mod) {
    x -= mod;
  }
  return x;
}
const int L = 20;
const int C = 1 << L;
int cnt[C + 1], cnt2[C + 1];
int pw2[N + 1];
struct Solution : Data {
  void solve() {
    int mx = 0;
    for (int i = 0; i < n; ++i) {
      ++cnt[a[i]];
    }
    for (int i = 0; i < L; ++i) {
      for (int j = 0; j < C; ++j) {
        if (j & (1 << i)) {
          cnt[j ^ (1 << i)] += cnt[j];
        }
      }
    }
    pw2[0] = 1;
    for (int i = 1; i <= n; ++i) {
      pw2[i] = sum(pw2[i - 1], pw2[i - 1]);
    }
    ans = 0;
    for (int i = 1; i < C; ++i) {
      long long e = -1;
      for (int j = 0; j < L; ++j) {
        if (i & (1 << j)) {
          e *= -1;
        }
      }
      ans += e * (pw2[cnt[i]] - 1);
      ans %= mod;
      ans += mod;
      ans %= mod;
    };
    ans = (pw2[n] - 1 - ans) % mod;
    ans += mod;
    ans %= mod;
  }
  Solution() {}
  Solution(Data d) : Data(d) {}
  void clear() {
    *this = Solution();
    memset(cnt, 0, sizeof(cnt));
    memset(cnt2, 0, sizeof(cnt2));
  }
};
Solution sol;
void test() {
  const int C = 1e6;
  long long cnt = 0;
  for (int i = 0; i <= C; ++i) {
    for (int j = i; j > 0; j = (j - 1) & i) {
      ++cnt;
    }
  };
}
int main() {
  sol.read();
  sol.solve();
  sol.write();
  return 0;
}