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 x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9')
    x = (x << 3) + (x << 1) + (c ^ '0'), c = getchar();
  return x * f;
}
const int N = 1 << 21, mod = 1e9 + 7, S = 1 << 20;
inline int inc(int x, int y) {
  x += y - mod;
  return x + (x >> 31 & mod);
}
inline void Inc(int &x, int y) {
  x += y - mod;
  x += (x >> 31 & mod);
}
inline void Del(int &x, int y) {
  x -= y;
  x += (x >> 31 & mod);
}
int n, f[N], cnt[N], bit[N], ans;
int main() {
  n = read();
  bit[0] = 1;
  for (int i = 1; i <= n; i++) bit[i] = inc(bit[i - 1], bit[i - 1]);
  for (int i = 1; i <= n; i++) f[read()]++;
  for (int i = 0; i < 20; i++)
    for (int s = 0; s < S; s++)
      if (s >> i & 1) f[s ^ (1 << i)] += f[s];
  ans = bit[n] - 1;
  for (int i = 1; i < S; i++) {
    cnt[i] = cnt[i >> 1] + (i & 1);
    (cnt[i] & 1) ? Del(ans, bit[f[i]] - 1) : Inc(ans, bit[f[i]] - 1);
  }
  printf("%d\n", ans);
  return 0;
}