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 inf = 1e9, mod = 1e9 + 7;
const int N = 2e6 + 5;
int P[N], n, a[N], i, j, k, m, dp[N];
long long ans;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    dp[a[i]]++;
  }
  P[0] = 1;
  for (int i = 1; i <= n; i++) P[i] = P[i - 1] * 2 % mod;
  for (int j = 0; j <= 19; j++)
    for (int i = (1 << 20) - 1; i >= 0; i--)
      if ((i & (1 << j))) dp[i - (1 << j)] += dp[i];
  for (int i = (1 << 20) - 1; i >= 0; i--) {
    int t = __builtin_popcount(i);
    if (t % 2)
      ans = (ans - P[dp[i]] + 1 + mod) % mod;
    else
      ans = (ans + P[dp[i]] - 1) % mod;
  }
  cout << ans << '\n';
  return 0;
}