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;
const double PI = acos(-1);
const double eps = 1e-9;
const int inf = 2000000000;
const long long infLL = 9000000000000000000;
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
inline long long gcd(long long a, long long b) {
  return b == 0 ? a : gcd(b, a % b);
}
inline long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
int dx[] = {0, 0, +1, -1, +1, +1, -1, -1};
int dy[] = {+1, -1, 0, 0, +1, -1, +1, -1};
inline void normal(long long &a) {
  a %= 1000000007;
  (a < 0) && (a += 1000000007);
}
inline long long modMul(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  return (a * b) % 1000000007;
}
inline long long modAdd(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  return (a + b) % 1000000007;
}
inline long long modSub(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  a -= b;
  normal(a);
  return a;
}
inline long long modPow(long long b, long long p) {
  long long r = 1;
  while (p) {
    if (p & 1) r = modMul(r, b);
    b = modMul(b, b);
    p >>= 1;
  }
  return r;
}
inline long long modInverse(long long a) { return modPow(a, 1000000007 - 2); }
inline long long modDiv(long long a, long long b) {
  return modMul(a, modInverse(b));
}
template <typename first, typename second>
ostream &operator<<(ostream &os, const pair<first, second> &p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << "{";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "}";
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename T>
ostream &operator<<(ostream &os, const multiset<T> &v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename first, typename second>
ostream &operator<<(ostream &os, const map<first, second> &v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << it->first << " = " << it->second;
  }
  return os << "]";
}
void faltu() { cerr << '\n'; }
template <typename T>
void faltu(T a[], int n) {
  for (int i = 0; i < n; ++i) cerr << a[i] << ' ';
  cerr << '\n';
}
template <typename T, typename... hello>
void faltu(T arg, const hello &...rest) {
  cerr << arg << ' ';
  faltu(rest...);
}
const int k = 20;
const int mx = 1e6 + 123;
long long dp[1 << k];
int a[mx];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int n, z = 0;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    dp[a[i]]++;
  }
  for (int i = 0; i < k; i++) {
    for (int mask = 0; mask < (1 << k); mask++) {
      if ((mask & (1 << i)) == 0) dp[mask] += dp[mask ^ (1 << i)];
    }
  }
  long long ans = 0;
  for (int mask = 0; mask < (1 << k); mask++) {
    int cnt = 0;
    for (int i = 0; i < k; i++) cnt += ((mask & (1 << i)) > 0);
    ans = modAdd(ans, modMul(modPow(-1, cnt), modSub(modPow(2, dp[mask]), 1)));
  }
  cout << ans;
  return 0;
}