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 = 1e9 + 7, mxN = 1e6 + 3;
long long n, dp[(1 << 20)];
const int k = 20, e = (1 << 20);
long long pc(int x) {
  long long res = 0;
  while (x > 0) res += (x & 1), x /= 2;
  return res;
}
long long po(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y & 1) res = res * x % mod;
    x = x * x % mod;
    y /= 2;
  }
  return res;
}
int main() {
  cin >> n;
  for (int i = 0, x; i < n; i++) {
    cin >> x;
    dp[x]++;
  }
  for (int i = 0; i < k; i++)
    for (long long j = 0; j < e; j++) {
      int g = j ^ (1 << i);
      if (j & (1 << i)) dp[g] += dp[j];
    }
  long long ans = 0;
  for (int i = 0; i < e; i++) {
    long long cnt = po(2, dp[i]) - 1;
    ans = (ans + ((pc(i) % 2) ? -cnt : cnt)) % mod;
  }
  cout << (ans + mod) % mod;
}