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;
void read(int &x) {
  char ch = getchar();
  int f = 1;
  while ((ch != '-') && (ch < '0' || ch > '9')) ch = getchar();
  if (ch == '-')
    f = -1, x = 0;
  else
    x = ch - 48;
  ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - 48, ch = getchar();
  x *= f;
}
void read(long long &x) {
  char ch = getchar();
  int f = 1;
  while ((ch != '-') && (ch < '0' || ch > '9')) ch = getchar();
  if (ch == '-')
    f = -1, x = 0;
  else
    x = ch - 48;
  ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10LL + ch - 48, ch = getchar();
  x *= f;
}
void _print(const int &x) {
  if (x < 10)
    putchar(48 + x);
  else
    _print(x / 10), _print(x % 10);
}
void _print(const long long &x) {
  if (x < 10)
    putchar(48 + x);
  else
    _print(x / 10), _print(x % 10);
}
void print(const int &x) {
  if (x < 0)
    putchar('-'), _print(-x);
  else
    _print(x);
}
void print(const long long &x) {
  if (x < 0)
    putchar('-'), _print(-x);
  else
    _print(x);
}
void println(const int &x) {
  print(x);
  putchar('\n');
}
void println(const long long &x) {
  print(x);
  putchar('\n');
}
int n, x, a[1 << 20], p2[1 << 20];
int main() {
  read(n);
  p2[0] = 1;
  for (int i = (1); i <= (n); i++)
    read(x), a[x]++, p2[i] = 1ll * p2[i - 1] * 2 % 1000000007;
  for (int i = (0); i <= (19); i++)
    for (int j = (0); j <= ((1 << 20) - 1); j++)
      if (!(j & 1 << i)) a[j] += a[j + (1 << i)];
  int ans = 0;
  for (int i = (0); i <= ((1 << 20) - 1); i++)
    if (__builtin_popcount(i) & 1)
      ans = (ans - p2[a[i]]) % 1000000007;
    else
      ans = (ans + p2[a[i]]) % 1000000007;
  println((ans + 1000000007) % 1000000007);
  return 0;
}