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 N = (1 << 20) + 10;
const int MOD = 1000000007;
int num[N], n;
long long pow2[N];
void init() {
  pow2[0] = 1;
  for (int i = 1; i < N; ++i) {
    pow2[i] = pow2[i - 1] * 2;
    if (pow2[i] >= MOD) pow2[i] -= MOD;
  }
}
int main() {
  init();
  scanf("%d", &n);
  int x;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &x);
    ++num[x];
  }
  long long ans = 0;
  for (int i = 0; i < 20; ++i) {
    for (int j = (1 << 20) - 1; j; --j) {
      if (j & (1 << i)) num[j - (1 << i)] += num[j];
    }
  }
  for (int i = (1 << 20) - 1; i; --i) {
    int k = __builtin_popcount(i);
    if (k % 2)
      ans += pow2[num[i]];
    else
      ans -= pow2[num[i]];
  }
  ans = pow2[n] - ans;
  ans %= MOD;
  if (ans < 0) ans += MOD;
  cout << ans << endl;
  return 0;
}