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 + 5, MOD = 1e9 + 7, LOG = 21;
int n, a[N], dp[2 * N][LOG];
long long jv;
long long power(long long a, long long b) {
  long long res = 1;
  for (; b; b /= 2) {
    if (b % 2) {
      res *= a;
      res %= MOD;
    }
    a *= a;
    a %= MOD;
  }
  return res;
}
void read_input() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    dp[a[i]][0]++;
  }
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  read_input();
  for (int j = 1; j < LOG; j++) {
    for (int i = 0; i < N; i++) {
      dp[i][j] = dp[i][j - 1];
      if ((i & (1 << (j - 1))) == 0) {
        dp[i][j] += dp[i + (1 << (j - 1))][j - 1];
      }
      dp[i][j] %= MOD;
    }
  }
  for (int i = 0; i < N; i++) {
    if (__builtin_popcount(i) % 2) {
      jv -= power(2, dp[i][LOG - 1]) - 1;
    } else {
      jv += power(2, dp[i][LOG - 1]) - 1;
    }
    jv %= MOD;
  }
  cout << (jv + MOD) % MOD;
  return 0;
}