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>
inline void Int(T &n) {
  n = 0;
  int f = 1;
  int ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (; isdigit(ch); ch = getchar()) n = (n << 3) + (n << 1) + ch - '0';
  n = n * f;
}
template <typename T>
T gcd(T a, T b) {
  return !b ? a : gcd(b, a % b);
}
template <typename T>
inline void umin(T &a, T b) {
  a = a < b ? a : b;
}
template <typename T>
inline void umax(T &a, T b) {
  a = a > b ? a : b;
}
template <typename T, typename W>
inline void Int(T &x, W &y) {
  Int(x), Int(y);
}
template <typename T, typename W, typename Q>
inline void Int(T &x, W &y, Q &z) {
  Int(x, y), Int(z);
}
const int N = 1e6 + 7;
const int inf = 1e9 + 7;
const int magic = 20;
int solve() {
  int n;
  Int(n);
  vector<int> dp(1 << magic);
  for (int i = 0; i < n; ++i) {
    int x;
    Int(x);
    ++dp[x];
  }
  for (int i = 0; i < magic; ++i) {
    for (int mask = 0; mask < (1 << magic); ++mask) {
      if ((mask >> i & 1) == 0) {
        dp[mask] += dp[mask | (1 << i)];
      }
    }
  }
  vector<long long> pw(n + 1);
  pw[0] = 1;
  for (int i = 1; i <= n; ++i) {
    pw[i] = (pw[i - 1] * 2) % 1000000007LL;
  }
  long long res = 0;
  for (int mask = 0; mask < (1 << magic); ++mask) {
    int x = (__builtin_popcount(mask) & 1) ? -1 : 1;
    res = (res + pw[dp[mask]] * x + 1000000007LL) % 1000000007LL;
  }
  printf("%lld\n", res);
  return 0;
}
int main() {
  int tests = 1, CaseNo = 0;
  while (tests--) {
    solve();
  }
  return 0;
}