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>
const int N = (int)1e6 + 7;
const int inf = (int)1e9 + 7;
const int mod = (int)1e9 + 7;
const long long linf = (long long)1e18 + 7;
const int dx[] = {-1, 0, 1, 0, 1, -1, -1, 1};
const int dy[] = {0, 1, 0, -1, 1, -1, 1, -1};
using namespace std;
void add(int &x, int y) {
  x += y;
  if (x >= mod)
    x -= mod;
  else if (x < 0)
    x += mod;
}
int sum(int x, int y) {
  add(x, y);
  return x;
}
int bp(int x, int y) {
  int res = 1;
  while (y) {
    if (y & 1) res = res * 1ll * x % mod;
    x = x * 1ll * x % mod, y >>= 1;
  }
  return res;
}
int n;
int dp[1 << 20];
int main() {
  ios_base ::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = (1); i <= (n); i++) {
    int x;
    cin >> x;
    dp[x]++;
  }
  for (int i = (0); i <= (19); i++) {
    for (int mask = ((1 << 20) - 1); mask >= (0); mask--) {
      if (mask & (1 << i)) dp[mask ^ (1 << i)] += dp[mask];
    }
  }
  int ans = 0;
  for (int mask = (0); mask <= ((1 << 20) - 1); mask++) {
    if (__builtin_popcount(mask) & 1)
      add(ans, -bp(2, dp[mask]));
    else
      add(ans, bp(2, dp[mask]));
  }
  cout << ans;
  exit(0);
}