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 EPS = 1e-6;
template <class T, class U>
ostream& operator<<(ostream& o, const pair<T, U>& v) {
  return o << "(" << v.first << ", " << v.second << ")";
}
template <class T>
ostream& operator<<(ostream& o, const vector<T>& v) {
  o << "{";
  for (int i = 0; i < (int)((int)((v).size())); ++i)
    o << (i ? ", " : "") << v[i];
  return o << "}";
}
void pkuassert(bool t) { t = 1 / t; };
int dx[] = {0, 1, 0, -1, 1, 1, -1, -1};
int dy[] = {1, 0, -1, 0, -1, 1, 1, -1};
int s2i(string& a) {
  stringstream ss(a);
  int r;
  ss >> r;
  return r;
}
int n;
int in[1 << 20];
int dp[21][1 << 20];
const long long mod = 1e9 + 7;
long long modpow(long long p, long long n, long long mod) {
  long long ret = 1;
  while (n) {
    if (n & 1) ret = (ret * p) % mod;
    p = p * p % mod;
    n >>= 1;
  }
  return ret;
}
const int bit = 20;
int main(int argc, char* argv[]) {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", in + i);
    ++dp[0][in[i]];
  }
  for (int i = 0; i < bit; ++i) {
    for (int j = 0; j < (int)(1 << bit); ++j) {
      if (j & (1 << i)) dp[i + 1][j ^ (1 << i)] += dp[i][j];
      dp[i + 1][j] += dp[i][j];
    }
  }
  long long ans = 0;
  for (int i = 0; i < (1 << bit); ++i) {
    if (__builtin_popcountll(i) % 2)
      ans -= modpow(2, dp[bit][i], mod);
    else
      ans += modpow(2, dp[bit][i], mod);
  }
  cout << (ans % mod + mod) % mod << endl;
  return 0;
}