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 long long inf = 1000000007;
int cnt[1024 * 1024 + 10][21];
int one[1024 * 1024 + 10];
void init() {
  vector<int> vt;
  for (int i = 1000000; i > 0; --i) {
    for (int j = 2; j < 21; ++j) cnt[i][j] /= j;
    vt.clear();
    for (int j = 0; j < 20; ++j)
      if (i & (1 << j)) one[i]++, vt.push_back(1 << j);
    for (int j = 0; j < 20; ++j)
      if (cnt[i][j])
        for (int k = 0; k < vt.size(); ++k) cnt[i ^ vt[k]][j + 1] += cnt[i][j];
  }
}
int prm[1024 * 1024 + 10];
int main() {
  prm[0] = 1;
  for (int i = 1; i < 1000010; ++i) prm[i] = prm[i - 1] * 2 % inf;
  int n, x;
  scanf("%d", &n);
  long long res = prm[n] - 1;
  while (n--) {
    scanf("%d", &x);
    cnt[x][0]++;
  }
  init();
  for (int i = 1; i <= 1000000; ++i) {
    int tmp = cnt[i][0];
    for (int j = 1; j <= 20; ++j) tmp += cnt[i][j];
    if (one[i] & 1)
      res = res - prm[tmp] + 1;
    else
      res = res + prm[tmp] - 1;
    if (res < 0)
      res += inf;
    else if (res >= inf)
      res -= inf;
  }
  cout << res << endl;
  return 0;
}