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;
int main() {
  int n;
  scanf("%d", &n);
  vector<vector<int>> dp((1 << 20), vector<int>(21, 0));
  for (int i = 0; i < n; i++) {
    int a;
    scanf("%d", &a);
    dp[a][0]++;
  }
  const long long mod = 1000000007;
  vector<long long> p(n + 1, 1);
  for (int i = 1; i <= n; i++) p[i] = (p[i - 1] * 2) % mod;
  for (int i = (1 << 20) - 1; i >= 0; i--) {
    for (int j = 1; j <= 20; j++) {
      dp[i][j] = dp[i][j - 1];
      if ((i & (1 << (j - 1))) == 0) dp[i][j] += dp[i + (1 << (j - 1))][j];
    }
  }
  long long res = 0;
  for (int i = 0; i < (1 << 20); i++) {
    int c = __builtin_popcount(i);
    if (c & 1)
      res = (res - p[dp[i][20]] + mod) % mod;
    else
      res = (res + p[dp[i][20]]) % mod;
  }
  printf("%I64d\n", res);
  return 0;
}