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 double eps = 1e-9;
const int maxmask = 1000010;
const int mod = 1000000007;
int cnt[maxmask], pw[maxmask];
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    int ai;
    scanf("%d", &ai);
    cnt[ai]++;
  }
  pw[0] = 1;
  for (int i = 1; i <= n; ++i) {
    pw[i] = (pw[i - 1] * 2) % mod;
  }
  for (int bit = 1; bit < maxmask; bit <<= 1) {
    for (int mask = 0; mask < maxmask; ++mask) {
      if (mask & bit) cnt[mask ^ bit] += cnt[mask];
    }
  }
  int res = pw[n] - 1;
  for (int i = 1; i < maxmask; ++i) {
    int bits = 0;
    for (int bit = 1; bit <= i; bit <<= 1) {
      if (i & bit) ++bits;
    }
    int add = pw[cnt[i]] - 1;
    if (bits & 1)
      res -= add;
    else
      res += add;
    res = res % mod;
  }
  res = (res + mod) % mod;
  cout << res;
  return 0;
}