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 mod = 1e9 + 7, limit = (1 << 21);
const int maxn = 2300000;
int n, a[maxn], f[21][maxn], g[maxn];
long long ans;
long long qpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return res;
}
long long get(long long x) {
  long long res = 0;
  for (long long i = 0; i < 21; i++)
    if (x >> i & 1) res++;
  return res;
}
int main() {
  scanf("%d", &n);
  for (long long i = 1; i <= n; i++) {
    long long t;
    scanf("%lld", &t);
    f[0][t]++;
  }
  for (long long i = 1; i <= 21; i++)
    for (long long j = 0; j <= limit; j++) {
      f[i][j] = f[i - 1][j];
      if (!(j >> (i - 1) & 1)) (f[i][j] += f[i - 1][j | (1 << (i - 1))]) %= mod;
    }
  for (long long i = 0; i <= limit; i++) g[i] = f[21][i];
  for (long long i = 0; i < limit; i++) {
    long long temp = qpow(-1, get(i));
    temp = temp * (qpow(2, g[i]) - 1 + mod) % mod;
    ans = (ans + temp + mod) % mod;
  }
  printf("%lld\n", (ans + mod) % mod);
  return 0;
}