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 lg = 20;
const int maxn = 2e6 + 100;
const int mod = 1e9 + 7;
const int inf = 1e9;
int n, a, c[maxn][2], dp[maxn][2], p[maxn];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  p[0] = 1;
  for (int i = 1; i < maxn; i++) {
    p[i] = (p[i - 1] * 2) % mod;
  }
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a;
    c[a][0]++;
  }
  for (int j = 1; j <= lg; j++) {
    for (int i = 0; i < (1ll << lg); i++) {
      c[i][j % 2] = c[i][1 - j % 2];
      if (!((i >> (j - 1)) & 1))
        c[i][j % 2] += c[i + (1ll << (j - 1))][1 - j % 2];
      c[i][j % 2] %= mod;
    }
  }
  for (int i = 0; i < (1ll << lg); i++) {
    dp[i][0] = p[c[i][lg % 2]] - 1;
    if (__builtin_popcount(i) % 2) dp[i][0] *= -1;
  }
  for (int j = 1; j <= lg; j++) {
    for (int i = 0; i < (1ll << lg); i++) {
      dp[i][j % 2] = dp[i][1 - j % 2];
      if (!((i >> (j - 1)) & 1))
        dp[i][j % 2] += dp[i + (1ll << (j - 1))][1 - j % 2];
      dp[i][j % 2] %= mod;
    }
  }
  cout << (dp[0][lg % 2] + mod) % mod;
  return 0;
}