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 MOD = (1e9) + 7;
int modPow(int b, int e) {
  if (e == 0)
    return 1;
  else if (e == 1)
    return b;
  return 1ll * modPow(1ll * b * b % MOD, e >> 1) * modPow(b, e & 1) % MOD;
}
int n, ans, f[1 << 20];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 0, x; i < n; i++) {
    cin >> x;
    f[x]++;
  }
  for (int j = 0; j < 20; j++)
    for (int mask = (1 << 20) - 1; mask >= 0; mask--) {
      if (!(mask >> j & 1)) f[mask] = (f[mask] + f[mask | 1 << j]) % MOD;
    }
  for (int mask = 0; mask < 1 << 20; mask++) {
    int bits = __builtin_popcount(mask);
    if (bits & 1)
      ans = (ans - modPow(2, f[mask])) % MOD;
    else
      ans = (ans + modPow(2, f[mask])) % MOD;
  }
  cout << (ans + MOD) % MOD;
}