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;
int n, m;
long long f[21][1 << 20];
long long pow2[1 << 20];
int main() {
  for (int i = 0; i < (int)(1 << 20); i++) {
    if (i == 0)
      pow2[0] = 1;
    else
      pow2[i] = (pow2[i - 1] * 2) % (1000000007L);
  }
  memset(f, 0, sizeof f);
  scanf("%d", &n);
  int x, maxi = 0;
  for (int i = 0; i < (int)n; i++) {
    scanf("%d", &x);
    maxi = max(maxi, x);
    f[0][x]++;
  }
  m = 0;
  while (maxi > 0) {
    maxi = maxi >> 1;
    m++;
  }
  for (int k = 1; k <= m; k++) {
    for (int i = 0; i < (1 << 20); i++) {
      if (((i >> k - 1) & 1)) {
        f[k][i] = f[k - 1][i];
      } else {
        f[k][i] = (f[k - 1][i] + f[k - 1][(i | ((1 << k - 1)))]) %
                  ((1000000007L) - 1);
      }
    }
  }
  long long sum = 0;
  for (int i = 0; i < (1 << 20); i++) {
    if (__builtin_popcount(i) % 2 == 0) {
      sum = (sum + (pow2[f[m][i]] - 1)) % (1000000007L);
    } else
      sum = (sum - (pow2[f[m][i]] - 1)) % (1000000007L);
  }
  sum = (sum + (1000000007L)) % (1000000007L);
  cout << sum << endl;
}