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>
long long read() {
  long long x = 0, f = 1;
  char ch;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch <= '9' && ch >= '0') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return f * x;
}
long long count(long long x) {
  long long ans = 0;
  for (; x; x -= (x & -x)) ans++;
  return ans;
}
long long fu(long long x) { return x & 1 ? -1 : 1; }
long long g[1050000], n, a, po[1050000];
long long ans;
int main() {
  n = read();
  po[0] = 1;
  for (long long i = 1; i <= n; i++) {
    a = read();
    g[a]++;
  }
  for (long long i = 0; i <= 20; i++) {
    for (long long j = 0; j <= 1000000; j++) {
      if (j & (1 << i)) g[j ^ (1 << i)] += g[j];
    }
  }
  for (long long i = 1; i <= 1000000; i++) po[i] = po[i - 1] * 2 % 1000000007;
  for (long long i = 0; i <= 1000000; i++) {
    ans = ((ans + fu(count(i)) * (po[g[i]] - 1)) % 1000000007 + 1000000007) %
          1000000007;
  }
  printf("%lld\n", ans);
  return 0;
}