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;
int n, a;
long long int f[21][1000005], ans;
long long int fastexp(long long int b, long long int e) {
  long long int res = 1;
  while (e) {
    if (e & 1) res = (res * b) % 1000000007;
    e >>= 1;
    b = (b * b) % 1000000007;
  }
  return res;
}
long long int dp(int k, int x) {
  if (x > 1000000) return 0;
  if (f[k][x] == -1) {
    f[k][x] = dp(k - 1, x);
    if (!(x & (1 << (k - 1))))
      f[k][x] = (f[k][x] + dp(k - 1, x + (1 << (k - 1)))) % 1000000007;
  }
  return f[k][x];
}
int main() {
  scanf("%d", &n);
  memset(f, -1, sizeof f);
  memset(f[0], 0, sizeof f[0]);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a);
    f[0][a]++;
  }
  ans = 0;
  for (int i = 0; i < (1 << 20); i++)
    if (__builtin_popcount(i) & 1)
      ans = (ans - fastexp((long long int)2, dp(20, i)) + 1000000007) %
            1000000007;
    else
      ans = (ans + fastexp((long long int)2, dp(20, i))) % 1000000007;
  printf("%lld\n", ans);
  return 0;
}