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 prime = 999983;
const int INF = 0x7FFFFFFF;
const long long INFF = 0x7FFFFFFFFFFFFFFF;
const double pi = acos(-1.0);
const double inf = 1e18;
const double eps = 1e-6;
const long long mod = 1e9 + 7;
long long qpow(long long a, long long b) {
  long long s = 1;
  while (b > 0) {
    if (b & 1) s = s * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return s;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
int dr[2][4] = {1, -1, 0, 0, 0, 0, -1, 1};
const int maxn = (1 << 20) + 10;
int F[maxn];
int A[maxn], B[maxn];
int pow2[maxn];
int ans[maxn];
int main(void) {
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &A[i]);
    B[A[i]]++;
  }
  int N = 20;
  pow2[0] = 1;
  for (int i = 1; i < (1 << N); ++i) pow2[i] = pow2[i - 1] * 2 % mod;
  for (int mask = 0; mask < (1 << N); ++mask) F[mask] = B[mask];
  for (int i = 0; i < N; ++i) {
    for (int mask = (1 << N) - 1; mask >= 0; --mask)
      if (mask & (1 << i)) F[mask ^ (1 << i)] += F[mask];
  }
  int ans = 0;
  for (int mask = 0; mask < (1 << N); ++mask) {
    int t = 1;
    if (__builtin_popcount(mask) & 1) t = -1;
    ans += 1ll * t * (pow2[F[mask]]) % mod;
    ans = (ans % mod + mod) % mod;
  }
  cout << ans << endl;
  return 0;
}