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 N = 1 << 20, mod = 1e9 + 7;
long long dp[2][N];
long long modexp(long long a, long long n) {
  if (n == 0) return 1;
  if (n == 1) return a;
  long long x = modexp(a, n / 2);
  x = (x * x) % mod;
  if (n & 1) x = (x * a) % mod;
  return x;
}
int32_t main() {
  ios ::sync_with_stdio(0);
  cin.tie(0);
  long long n;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    long long p;
    cin >> p;
    dp[1][p]++;
  }
  for (long long i = 19; i >= 0; i--) {
    long long cur = i & 1;
    for (long long mask = 0; mask < N; mask++) {
      dp[cur ^ 1][mask] = dp[cur][mask];
      if (mask & (1 << i)) {
        dp[cur ^ 1][mask ^ (1 << i)] += dp[cur][mask];
      }
    }
  }
  long long ans = 0;
  for (long long i = 0; i < N; i++) {
    if (__builtin_popcount(i) % 2) {
      ans -= modexp(2, dp[1][i]) - 1;
    } else {
      ans += modexp(2, dp[1][i]) - 1;
    }
    ans = (ans + mod) % mod;
  }
  cout << ans;
  return 0;
}