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;
extern inline char gc() {
  static char RR[23456], *S = RR + 23333, *T = RR + 23333;
  if (S == T) fread(RR, 1, 23333, stdin), S = RR;
  return *S++;
}
inline int read() {
  int p = 0, w = 1;
  char c = gc();
  while (c > '9' || c < '0') {
    if (c == '-') w = -1;
    c = gc();
  }
  while (c >= '0' && c <= '9') p = p * 10 + c - '0', c = gc();
  return p * w;
}
int n;
int f[2105000], bit[2105000], ans;
int main() {
  n = read();
  for (register int i = 1; i <= n; i++) f[read()]++;
  bit[0] = 1;
  for (register int i = 1; i <= 1000000; i++)
    bit[i] = 1ll * 2 * bit[i - 1] % 1000000007;
  for (register int i = 0; i <= 19; i++)
    for (register int j = 0; j <= 1000000; j++)
      if (!(j & bit[i])) f[j] += f[j ^ bit[i]];
  for (register int i = 0; i <= 1000000; i++) {
    int cnt = 1;
    for (register int j = 0; j <= 19; j++)
      if (i & bit[j]) cnt = -cnt;
    int tmp = cnt * (bit[f[i]] - 1) % 1000000007;
    ans = ((ans + tmp) % 1000000007 + 1000000007) % 1000000007;
  }
  printf("%d\n", ans);
  return 0;
}