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;
long long inline read() {
  long long num = 0, neg = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') neg = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    num = (num << 3) + (num << 1) + c - '0';
    c = getchar();
  }
  return num * neg;
}
const long long maxn = 1e6 + 10, mod = 1e9 + 7;
long long n, a[maxn], f[1 << 20], g[1 << 20], sum[1 << 20];
long long Pow(long long x, long long y) {
  long long ans = 1;
  while (y) {
    if (y & 1) ans = ans * x % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return ans;
}
long long Get(long long x) {
  long long num = 0;
  while (x) {
    if (x & 1) num++;
    x >>= 1;
  }
  return num;
}
signed main() {
  n = read();
  for (long long i = 1; i <= n; i++) a[i] = read(), f[a[i]]++;
  for (long long i = 0; i < 20; i++)
    for (long long j = (1 << 20) - 1; j >= 0; j--)
      if (!(j & (1 << i))) f[j] = f[j] + f[j | (1 << i)];
  for (long long i = 0; i < (1 << 20); i++)
    g[i] = ((Pow(2, f[i]) - 1) % mod + mod) % mod;
  for (long long i = 0; i < (1 << 20); i++) sum[i] = (Get(i) & 1) ? -1 : 1;
  long long ans = 0;
  for (long long i = 0; i < (1 << 20); i++)
    ans = (ans + (sum[i] * g[i] % mod + mod) % mod) % mod;
  cout << ans << endl;
  return 0;
}