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 long long read() {
  long long sum = 0, ff = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') ff = -1;
    ch = getchar();
  }
  while (isdigit(ch)) sum = sum * 10 + (ch ^ 48), ch = getchar();
  return sum * ff;
}
const long long N = 1e6 + 5;
const long long mo = 1e9 + 7;
long long n, m, ans, f[22], dp[N], g[N];
inline void Add(long long &x, long long y) {
  x += y;
  if (x >= mo) x -= mo;
}
signed main() {
  for (long long i = 1; g[0] = 1, i <= 1e6; i++) g[i] = (g[i - 1] << 1) % mo;
  n = read();
  for (long long i = 1; i <= n; i++) dp[read()]++;
  for (long long j = 0; j <= 21; j++)
    for (long long i = 1; i <= 1e6; i++)
      if ((1ll << j) & i) Add(dp[i ^ (1ll << j)], dp[i]);
  for (long long i = 0; i <= 1e6; i++) {
    long long gs = 0;
    for (long long j = 0; j <= 21; j++)
      if ((1ll << j) & i) gs++;
    Add(f[gs], g[dp[i]] - 1);
  }
  for (long long i = 0; i <= 21; i++)
    (i & 1) ? Add(ans, -f[i]) : Add(ans, f[i]);
  printf("%lld\n", (ans + mo) % mo);
  return 0;
}