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 T read(T& x) {
  x = 0;
  int f = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') f |= (ch == '-'), ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x = f ? -x : x;
}
const int N = 1e6 + 10;
const int P = 1e9 + 7;
int n, i, status, ans, a[N], f[(1 << 20) + 10];
int fexp(int a, int n) {
  int res = 1;
  while (n) {
    if (n & 1) res = 1LL * res * a % P;
    a = 1LL * a * a % P;
    n >>= 1;
  }
  return res;
}
inline void up(int& a, int b) {
  a += b;
  if (a >= P) a -= P;
}
int main() {
  read(n);
  for (i = 1; i <= n; i++) {
    read(a[i]);
    f[a[i]]++;
  }
  for (i = 0; i < 20; i++) {
    for (status = 0; status < (1 << 20); status++) {
      if (!(status & (1 << i))) f[status] += f[status | (1 << i)];
    }
  }
  for (status = 0; status < (1 << 20); status++) {
    int g = __builtin_popcount(status);
    int res = fexp(2, f[status]) - 1;
    if (res < 0) res += P;
    if (g & 1)
      up(ans, P - res);
    else
      up(ans, res);
  }
  printf("%d\n", ans);
  return 0;
}