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[4000010], F[4000010], limit, ans = 0;
void FWT_AND(int *A, int type) {
  for (register int mid = 1; mid < limit; mid <<= 1) {
    for (register int R = (mid << 1), j = 0; j < limit; j += R) {
      for (register int k = 0; k < mid; k++) {
        if (type)
          A[j + k] = (A[j + k] + A[j + k + mid]) % 1000000007;
        else
          A[j + k] = (A[j + k] - A[j + k + mid] + 1000000007) % 1000000007;
      }
    }
  }
}
int fpm(int x, int y) {
  int res = 1;
  while (y) {
    if (y & 1) res = 1ll * res * x % 1000000007;
    x = 1ll * x * x % 1000000007;
    y >>= 1;
  }
  return res;
}
int main() {
  scanf("%d", &n);
  limit = (1 << 20);
  for (register int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (register int i = 1; i <= n; ++i) F[a[i]]++;
  FWT_AND(F, 1);
  for (register int i = 0; i < limit; ++i) {
    int cnt = 0, x = i;
    while (x) {
      cnt += (x & 1);
      x >>= 1;
    }
    if (cnt & 1)
      ans = (ans - fpm(2, F[i]) + 1000000007) % 1000000007;
    else
      ans = (ans + fpm(2, F[i])) % 1000000007;
  }
  printf("%d\n", ans);
  return 0;
}