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 mod = 1e9 + 7;
const int max_n = 1e6 + 5;
int dp[max_n];
long long f[21], b[max_n];
int main() {
  int n;
  scanf("%d", &n);
  memset(dp, 0, sizeof dp);
  int mx = 0;
  b[0] = 1;
  for (int i = 1; i <= n; i++) b[i] = (b[i - 1] << 1) % mod;
  for (int i = 1; i <= n; i++) {
    int a_i;
    scanf("%d", &a_i);
    mx = max(mx, a_i);
    dp[a_i]++;
  }
  for (int j = 0; j <= 20; j++) {
    for (int i = 1; i <= mx; i++)
      if (dp[i]) {
        if (i & (1 << j)) {
          dp[i ^ (1 << j)] += dp[i];
        }
      }
  }
  for (int i = 0; i <= mx; i++) {
    int cnt = 0;
    for (int j = 0; j <= 20; j++) {
      if (i & (1 << j)) cnt++;
    }
    f[cnt] = (f[cnt] + (b[dp[i]] - 1) % mod) % mod;
  }
  long long sum = 0;
  for (int i = 0; i <= 20; i++) {
    if (i % 2 == 0)
      sum = (sum + f[i]) % mod;
    else
      sum = (sum - f[i] + mod) % mod;
  }
  printf("%lld\n", sum);
  return 0;
}