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 = 1000100;
const long long MOD = 1e9 + 7;
int n;
long long fac[MAXN], inv[MAXN];
long long quick_pow(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % MOD;
    a = a * a % MOD;
    b >>= 1;
  }
  return ret;
}
void init() {
  fac[0] = 1;
  for (int i = 1; i < MAXN; ++i) fac[i] = fac[i - 1] * i % MOD;
  inv[MAXN - 1] = quick_pow(fac[MAXN - 1], MOD - 2);
  for (int i = MAXN - 2; i >= 0; --i) inv[i] = inv[i + 1] * (i + 1) % MOD;
  assert(inv[0] == 1);
}
long long C(int n, int m) {
  if (m > n) return 0;
  return fac[n] * inv[m] % MOD * inv[n - m] % MOD;
}
long long dp[MAXN], f[MAXN];
int main() {
  init();
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    int x;
    scanf("%d", &x);
    ++dp[x];
  }
  for (int i = 0; i < 20; ++i) {
    for (int s = 0; s < MAXN; ++s)
      if (s & (1 << i)) dp[s ^ (1 << i)] += dp[s];
  }
  long long ans = quick_pow(2, n) - 1;
  f[0] = 0;
  for (int s = 1; s < MAXN; ++s) {
    f[s] = f[(s - 1) & s] + 1;
    int flag = (f[s] & 1) ? -1 : 1;
    ans = (ans + flag * (quick_pow(2, dp[s]) - 1)) % MOD;
  }
  ans = (ans % MOD + MOD) % MOD;
  cout << ans << endl;
}