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 N = 1e6 + 6, MOD = 1e9 + 7, B = 21;
int n, arr[N], dp[(1 << B)][B], ans = 0, pow2[N];
int M(int x) {
  while (x < 0) x += MOD;
  return x % MOD;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  pow2[0] = 1;
  for (int i = 1; i < N; i++) {
    pow2[i] = (pow2[i - 1] * 2) % MOD;
  }
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    dp[arr[i]][B - 1] += 1;
  }
  for (int i = B - 2; i >= 0; i--) {
    for (int j = (1 << B) - 1; j >= 0; j--) {
      if (j & (1 << i)) {
        dp[j][i] = dp[j][i + 1];
      } else {
        dp[j][i] = dp[j][i + 1] + dp[j ^ (1 << i)][i + 1];
      }
    }
  }
  for (int i = 0; i < (1 << B); i++) {
    int cnt = 0, multip = 1;
    for (int j = 0; j < B; j++) {
      if (i & (1 << j)) cnt++;
    }
    if (cnt % 2) multip *= (-1);
    ans += pow2[dp[i][0]] * multip;
    ans = M(ans);
  }
  cout << ans;
}