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 MAXN = 1048576 + 50;
const int mod = 1e9 + 7;
int n, f[MAXN], pw[MAXN];
void fwt(int *a, int len, int flg) {
  for (int i = 2; i <= len; i <<= 1)
    for (int j = 0; j < len; j += i)
      for (int k = j; k < j + i / 2; ++k)
        a[k] = (a[k] + 1ll * (flg == 1 ? 1 : mod - 1) * a[k + i / 2]) % mod;
}
int main() {
  scanf("%d", &n);
  int len = pw[0] = 1;
  for (int i = 1, x; i <= n; ++i) {
    scanf("%d", &x), ++f[x];
    while (len <= x) len <<= 1;
    pw[i] = pw[i - 1] * 2ll % mod;
  }
  fwt(f, len, 1);
  for (int i = 0; i < len; ++i) f[i] = pw[f[i]];
  fwt(f, len, -1);
  printf("%d\n", f[0] == pw[n] ? f[0] - 1 : f[0]);
}