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 int N = 1e6 + 5;
const int mod = 1e9 + 7;
int n;
int cnt[1 << 22], two[N];
template <typename T>
inline void chkmax(T &x, T y) {
  x = max(x, y);
}
template <typename T>
inline void chkmin(T &x, T y) {
  x = min(x, y);
}
template <typename T>
inline void read(T &x) {
  T f = 1;
  x = 0;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -f;
  for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
  x *= f;
}
inline void inc(int &x, int y) { x = x + y < mod ? x + y : x + y - mod; }
inline void dec(int &x, int y) { x = x - y >= 0 ? x - y : x - y + mod; }
int main() {
  read(n);
  for (int i = 1; i <= n; ++i) {
    int x;
    read(x);
    ++cnt[x];
  }
  two[0] = 1;
  for (int i = 1; i <= n; ++i) two[i] = 2ll * two[i - 1] % mod;
  for (int i = 0; i <= 20; ++i)
    for (int s = 0; s < (1 << 20); ++s)
      if (s >> i & 1) cnt[s ^ (1 << i)] += cnt[s];
  int ans = 0;
  for (int s = 0; s < (1 << 20); ++s)
    if (__builtin_popcount(s) & 1)
      dec(ans, (two[cnt[s]] - 1 + mod) % mod);
    else
      inc(ans, (two[cnt[s]] - 1 + mod) % mod);
  printf("%d\n", ans);
  return 0;
}