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 Mo = 1000000007;
int All, n, i, j, k;
int a[2000005], b[2000005], f[2000005];
long long two[2000005], ans, tmp;
int main() {
  All = (1 << 20) - 1;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d", &a[i]), b[a[i]]++;
  two[0] = 1;
  for (i = 1; i <= n; i++) two[i] = two[i - 1] * 2 % Mo;
  for (i = 0; i < 20; i++)
    for (j = 0; j <= All; j++)
      if (j & (1 << i)) b[j - (1 << i)] += b[j], f[j]++;
  for (i = 0; i <= All; i++) {
    tmp = (two[b[i]] + Mo - 1) % Mo;
    if (f[i] & 1)
      ans = (ans + Mo - tmp) % Mo;
    else
      ans = (ans + tmp) % Mo;
  }
  printf("%I64d\n", ans);
}