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 long long N = 2e6 + 10;
const long long p = 1e9 + 7;
long long n, a[N], f[N], ans;
long long get(long long x) {
  long long sum = 0;
  while (x) x -= (x & -x), sum++;
  return sum;
}
long long POW(long long a, long long b) {
  if (!b) return 1;
  if (b == 1) return a;
  long long sum = POW(a, b / 2);
  if (b % 2) return sum * sum % p * a % p;
  return sum * sum % p;
}
signed main() {
  scanf("%lld", &n);
  for (long long i = 1; i <= n; i++) scanf("%lld", &a[i]), f[a[i]]++;
  for (long long j = 0; j < 20; j++)
    for (long long i = (1 << 20) - 1; i >= 0; i--)
      if ((i & (1 << j)) == 0) f[i] += f[i + (1 << j)];
  for (long long i = 0; i < (1 << 20); i++)
    ans += (get(i) % 2 ? -1 : 1) * (POW(2, f[i]) - 1 + p) % p, ans += p,
        ans %= p;
  printf("%lld", ans);
  return 0;
}