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>
const int num = 1000000, mo = 1e9 + 7;
int a[num + 5], f[num + 5], i, j, n, x, ans = 0;
int main() {
  scanf("%d\n", &n);
  for (i = 1; i <= n; ++i) scanf("%d", &x), f[x]++;
  for (a[0] = i = 1; i <= n; ++i) a[i] = (a[i - 1] << 1) % mo;
  for (j = 0; j < 20; ++j)
    for (i = num; i >= 0; --i)
      if (i & (1 << j)) (f[1 << j ^ i] += f[i]) %= mo;
  for (i = num; i >= 0; --i) {
    x = 1;
    for (j = 0; j < 20; ++j)
      if (i & (1 << j)) x = -x;
    (ans += 1ll * x * (a[f[i]] - 1)) %= mo;
  }
  printf("%d\n", (ans + mo) % mo);
}