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;
const int N = 2e6 + 9;
int n, f[N], a[N];
long long qpow(long long x, long long y) {
  long long t = 1;
  while (y) {
    if (y & 1) (t *= x) %= mod;
    y >>= 1, (x *= x) %= mod;
  }
  return t;
}
long long g[N], t[N];
int lowbit(int x) { return x & (-x); }
int main() {
  scanf("%d", &n);
  int M = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    f[a[i]]++;
    M = max(M, a[i]);
  }
  for (int i = 0; i < 20; i++)
    for (int j = M - (1 << i); j >= 0; j--)
      if (!(j & (1 << i))) f[j] += f[j | (1 << i)];
  for (int i = (1 << 20) - 1; i >= 0; i--)
    g[i] = ((qpow(2, f[i]) - 1) % mod + mod) % mod;
  t[0] = 0;
  for (int i = 1; i < (1 << 20); i++) t[i] = t[i - lowbit(i)] + 1;
  long long ans = qpow(2, n);
  for (int i = 1; i < (1 << 20); i++) {
    long long k = (t[i] % 2 == 1) ? (-1) : 1;
    ans = ((ans + k * g[i] % mod) % mod + mod) % mod;
  }
  printf("%lld\n", ans - 1);
  return 0;
}