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 N = 20;
const long long K = 21;
const long long maxk = 1e6 + 6;
const long long maxn = 1e6;
const long long mod = 1e9 + 7;
const long long p2N = ((1 << N) - 1);
long long dp[p2N + 1], n, arr[maxk], p2[maxk], ans, tmp;
long long add(long long a, long long b, long long md) {
  long long ret = a + b;
  if (ret >= md) ret -= md;
  return ret;
}
long long dif(long long a, long long b, long long md) {
  long long ret = a - b;
  if (ret < 0) ret += md;
  return ret;
}
int main() {
  for (int i = 0; i < maxk; i++) p2[i] = (i == 0) ? 1 : (p2[i - 1] * 2) % mod;
  scanf("%lld", &n);
  for (int i = 0; i < n; i++) {
    scanf("%lld", &arr[i]);
    dp[p2N ^ arr[i]]++;
  }
  for (int i = 0; i < N; i++)
    for (int j = 0; j <= p2N; j++)
      if (!(j & (1 << i))) dp[j + (1 << i)] += dp[j];
  for (int i = 0; i <= p2N; i++) {
    if (__builtin_popcount(i) % 2 & 1)
      ans = dif(ans, p2[dp[i]], mod);
    else
      ans = add(ans, p2[dp[i]], mod);
  }
  printf("%lld\n", ans);
  return 0;
}