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 mk = 1 << 21;
const long long logn = 29;
const long long mod = 1e9 + 7;
long long n, num[mk];
long long f[mk], qpow[mk], g[mk], ans;
signed main() {
  cin >> n;
  int mx = 1000000;
  qpow[0] = 1;
  for (int a, i = 1; i <= n; i++) {
    scanf("%d", &a), f[a]++;
    qpow[i] = (qpow[i - 1] << 1) % mod;
  }
  for (int i = 0; i <= 20; i++) {
    for (int j = 1; j <= mx; j++)
      if (j & (1 << i)) f[j ^ (1 << i)] += f[j];
  }
  for (int i = 0; i <= mx; i++) {
    num[i] = num[i >> 1] + (i & 1);
    g[num[i]] = (g[num[i]] + qpow[f[i]] - 1) % mod;
  }
  for (int i = 0; i <= 20; i++) {
    if (i & 1)
      ans = (ans - g[i] + mod) % mod;
    else
      ans = (ans + g[i]) % mod;
  }
  cout << ans;
}