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>
void read(T &first) {
  int f = 1;
  first = 0;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    first = (first << 3) + (first << 1) + c - '0';
    c = getchar();
  }
  first *= f;
  return;
}
void write(register int first) {
  if (first < 0) {
    first = ~first + 1;
    putchar('-');
  }
  if (first > 9) write(first / 10);
  putchar(first % 10 + '0');
  return;
}
int n;
long long f[(1 << 20)], ans;
long long power(long long first, long long second) {
  long long ret = 1;
  for (; second; second >>= 1, first = first * first % (int)(1e9 + 7))
    ret = ret * (second & 1 ? first : 1) % (int)(1e9 + 7);
  return ret;
}
int count(int first) {
  int ret = 0;
  for (; first; first >>= 1) ret += first & 1;
  return ret;
}
int main() {
  read(n);
  for (register int(i) = (1); (i) <= (n); i++) {
    int first;
    read(first);
    f[first]++;
  }
  for (register int(i) = (0); (i) <= (19); i++)
    for (register int(j) = ((1 << 20) - 1); (j) >= (0); j--)
      if (!(j & (1 << i))) (f[j] += f[j | (1 << i)]) %= (int)(1e9 + 7);
  for (register int(i) = (0); (i) <= ((1 << 20) - 1); i++)
    (ans += (power(2, f[i]) - 1) * (count(i) & 1 ? -1 : 1)) %= (int)(1e9 + 7);
  write((ans + (int)(1e9 + 7)) % (int)(1e9 + 7));
  return 0;
}