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 N = 1e6 + 5;
const long long mo = 1e9 + 7;
long long f[N + 1], g[N + 1], b[N + 1], ans;
int n, a[N], tot;
int main() {
  scanf("%d", &n);
  for (int(i) = (1); (i) <= (n); (i)++) scanf("%d", &a[i]), f[a[i]]++;
  b[0] = 1;
  for (int(i) = (1); (i) <= (N); (i)++) b[i] = b[i - 1] * 2ll % mo;
  for (int(j) = (0); (j) <= (20); (j)++) {
    for (int(i) = (0); (i) <= (N); (i)++)
      if (i & b[j]) f[i ^ b[j]] += f[i];
  }
  for (int(i) = (0); (i) <= (N); (i)++) {
    tot = 0;
    for (int(j) = (0); (j) <= (20); (j)++)
      if (i & b[j]) tot++;
    g[tot] = (g[tot] + b[f[i]] - 1) % mo;
  }
  for (int(i) = (0); (i) <= (20); (i)++) {
    if (i % 2 == 0)
      ans = (ans + g[i]) % mo;
    else
      ans = (ans - g[i]) % mo;
  }
  printf("%lld", (ans % mo + mo) % mo);
  return 0;
}