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 a[1 << 20][21];
int d[1 << 20];
int pow1[1 << 20];
int main() {
  int i, j, n, ans, cnt, x, k;
  pow1[0] = 1;
  for (i = 1; i < (1 << 20); i++) {
    pow1[i] = pow1[i - 1] + pow1[i - 1];
    if (pow1[i] >= 1000000007) pow1[i] -= 1000000007;
  }
  for (i = 0; i < (1 << 20); i++) {
    pow1[i]--;
    if (pow1[i] < 0) pow1[i] += 1000000007;
  }
  scanf("%d", &n);
  memset(a, 0, sizeof(a));
  for (i = 0; i < n; i++) {
    scanf("%d", &x);
    a[x][20]++;
  }
  memset(d, 0, sizeof(d));
  for (i = (1 << 20) - 1; i >= 0; i--)
    for (j = 0; j < 21; j++)
      if (a[i][j] > 0) {
        d[i] += a[i][j];
        for (k = j - 1; k >= 0; k--)
          if (((1 << k) & i) != 0) a[i ^ (1 << k)][k] += a[i][j];
      }
  ans = 0;
  for (i = 1; i < (1 << 20); i++) {
    cnt = 0;
    for (j = 0; j < 20; j++)
      if (((1 << j) & i) != 0) cnt++;
    if (cnt % 2 == 0) {
      ans -= pow1[d[i]];
      if (ans < 0) ans += 1000000007;
    } else {
      ans += pow1[d[i]];
      if (ans >= 1000000007) ans -= 1000000007;
    }
  }
  ans = pow1[n] - ans;
  if (ans < 0) ans += 1000000007;
  printf("%d\n", ans);
  return 0;
}