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 M = 1000000007;
const int N = 2000010;
int n, co[N], f[22][N], p2[N], m, k;
int parity(int n) {
  int num = 0;
  if (n) do
      num ^= 1;
    while (n &= n - 1);
  return num;
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n;
  for (int i = 1, x; i <= n; ++i) {
    cin >> x;
    co[x]++;
    m = max(m, x);
  }
  k = 1;
  while ((1 << k) <= m) ++k;
  p2[0] = 1;
  for (int i = 1; i <= n; ++i) p2[i] = (2 * p2[i - 1]) % M;
  if (n == N - 10) {
    cout << k << ' ' << m << '\n';
  }
  int ans = 0;
  for (int c = (1 << k) - 1; c >= 0; --c) {
    if (c & 1)
      f[0][c] = co[c];
    else
      f[0][c] = co[c] + co[c ^ 1];
    for (int i = 1; i < k; ++i)
      if (c & (1 << i))
        f[i][c] = f[i - 1][c];
      else
        f[i][c] = (f[i - 1][c] + f[i - 1][c ^ (1 << i)]) % M;
    int sgn = parity(c);
    if (sgn % 2)
      sgn = -1;
    else
      sgn = 1;
    ans = (1LL * ans + sgn * p2[f[k - 1][c]]) % M;
    if (ans < 0) ans += M;
  }
  cout << ans << '\n';
}