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 = (int)1e9 + 7;
long long dp[1 << 20];
long long exp(long long a, int b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return ans;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int a;
    scanf("%d", &a);
    dp[a]++;
  }
  for (int i = 0; i < 20; i++) {
    for (int mask = 0; mask < (1 << 20); mask++) {
      if (mask & (1 << i)) {
        dp[mask ^ (1 << i)] += dp[mask];
        dp[mask] %= mod;
      }
    }
  }
  long long ans = 0;
  for (int mask = 0; mask < (1 << 20); mask++) {
    if (dp[mask] == 0) continue;
    if (__builtin_popcount(mask) & 1)
      ans = (ans - (exp(2LL, dp[mask]) - 1) + mod) % mod;
    else
      ans = (ans + exp(2LL, dp[mask]) - 1) % mod;
  }
  printf("%lld\n", ans);
  return 0;
}