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;
long long gcd(long long u, long long v) {
  while (v != 0) {
    long long r = u % v;
    u = v;
    v = r;
  }
  return u;
}
const int MOD = 1000000007;
int n;
int a[1000000];
int p2[1000000];
int cnt[1 << 20];
int h[1 << 20];
void calc(int k, int y) {
  if (k == 0) {
    h[y] = cnt[y];
    return;
  }
  calc(k - 1, y);
  calc(k - 1, y + (1 << (k - 1)));
  for (int i = (0); i < (1 << (k - 1)); ++i)
    h[y + i] += h[y + i + (1 << (k - 1))];
}
void run() {
  scanf("%d", &n);
  for (int i = (0); i < (n); ++i) scanf("%d", &a[i]);
  memset(cnt, 0, sizeof(cnt));
  for (int i = (0); i < (n); ++i) ++cnt[a[i]];
  p2[0] = 1;
  for (int i = (1); i <= (n); ++i) {
    p2[i] = p2[i - 1] + p2[i - 1];
    if (p2[i] >= MOD) p2[i] -= MOD;
  }
  calc(20, 0);
  int ret = 0;
  for (int i = (0); i < (1 << 20); ++i) {
    int g = h[i];
    int f = p2[g] - 1;
    if (f < 0) f += MOD;
    int sign = +1;
    for (int j = (0); j < (20); ++j)
      if (i & (1 << j)) sign = -sign;
    if (sign == +1) {
      ret = ret + f;
      if (ret >= MOD) ret -= MOD;
    } else {
      ret = ret - f;
      if (ret < 0) ret += MOD;
    }
  }
  printf("%d\n", ret);
}
int main() {
  run();
  return 0;
}