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;
inline int read() {
  int ra, fh;
  char rx;
  rx = getchar(), ra = 0, fh = 1;
  while ((rx < '0' || rx > '9') && rx != '-') rx = getchar();
  if (rx == '-') fh = -1, rx = getchar();
  while (rx >= '0' && rx <= '9') ra *= 10, ra += rx - 48, rx = getchar();
  return ra * fh;
}
int n, x, dp[3000000], g[3000000];
long long f[3000000];
long long ans = 0;
int main() {
  scanf("%d", &n);
  int mx = (1 << 20) - 1;
  for (int i = 1; i <= n; i++) {
    x = read();
    dp[x]++;
  }
  f[0] = 1;
  for (int i = 1; i <= n; i++) f[i] = (f[i - 1] << 1) % 1000000007;
  for (int i = 20; i >= 1; i--)
    for (int j = mx; j >= 0; j--)
      if (!(j & (1 << (i - 1)))) dp[j] += dp[j | (1 << (i - 1))];
  ans = f[n] - 1;
  for (int i = 1; i <= mx; i++) {
    g[i] = g[i >> 1] + (1 & i);
    if (g[i] & 1)
      ans = (ans - f[dp[i]] + 1 + 1000000007) % 1000000007;
    else
      ans = (ans + f[dp[i]] - 1 + 1000000007) % 1000000007;
  }
  printf("%lld\n", ans);
  return 0;
}