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 B = 1e9 + 7;
const int N = 1 << 20;
int n;
int f[21][N];
long long mypow(long long a, int b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a % B;
    a = a * a % B;
    b >>= 1;
  }
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = 0, a; i < n; ++i) {
    scanf("%d", &a);
    if (a & 1) {
      f[0][a]++;
      f[0][a ^ 1]++;
    } else {
      f[0][a]++;
    }
  }
  for (int i = 1; i <= 20; ++i) {
    for (int x = 0; x < N; ++x) {
      if (x >> i & 1) {
        f[i][x] = f[i - 1][x];
      } else {
        f[i][x] = f[i - 1][x] + f[i - 1][x + (1 << i)];
      }
    }
  }
  long long ans = 0;
  for (int x = 0; x < N; ++x) {
    if (__builtin_popcount(x) & 1) {
      ans = (ans + B - mypow(2, f[20][x])) % B;
    } else {
      ans = (ans + mypow(2, f[20][x])) % B;
    }
  }
  printf("%lld\n", ans);
  return 0;
}