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 maxN = 2e6 + 5;
const int maxLOG = 20;
const long long INF = 1e18;
const long long MOD = 1e9 + 7;
long long gcd(long long a, long long b) { return !b ? a : gcd(b, a % b); }
long long sq(long long x) { return (x * x) % MOD; }
long long modP(long long a, long long b) {
  return (!b ? 1 : (sq(modP(a, b / 2)) * (b % 2 ? a : 1)) % MOD);
}
long long sub[maxN], dp[maxN], a[maxN];
long long NT(long long x) { return (1 << maxLOG) - x - 1; }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    sub[NT(a[i])] += 1;
  }
  for (long long i = 0; i < maxLOG; i++) {
    for (long long j = 0; j < maxN; j++) {
      if (j & (1 << i)) {
        sub[j] += sub[j ^ (1 << i)];
        sub[j] %= MOD;
      }
    }
  }
  for (int i = 0; i < maxN; i++) {
    if (i < NT(i)) swap(sub[i], sub[NT(i)]);
  }
  long long res = 0;
  long long mx = *max_element(a, a + n);
  for (int i = 0; i <= mx; i++) {
    res += (__builtin_popcount(i) % 2 ? -1 : +1) * (modP(2, sub[i]) - 1);
    while (res < MOD) res += MOD;
    res %= MOD;
  }
  cout << res << endl;
  return 0;
}