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;
const int B = 20, T = (1 << B) + 1, MOD = (int)1e9 + 7;
int n, a[T], c[T], p2[T], dp[2][T];
inline int add(int u, int v) {
  u += v;
  return (u >= MOD) ? u - MOD : u;
}
void process() {
  for (int i = 0; i < T; i++) dp[1][i] = c[i];
  for (int i = 0; i < B; i++) {
    int s = i & 1;
    for (int j = 0; j < T - 1; j++) {
      int u = j & (1 << i);
      if (u)
        dp[s][j] = dp[s ^ 1][j];
      else {
        dp[s][j] = dp[s ^ 1][j] + dp[s ^ 1][j + (1 << i)];
      }
    }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  for (int i = 1; i <= n; i++) ++c[a[i]];
  p2[0] = 1;
  for (int i = 1; i < T; i++) p2[i] = add(p2[i - 1], p2[i - 1]);
  int ans = 1;
  process();
  for (int i = 0; i < T; i++) {
    int pc = __builtin_popcount(i);
    if (pc & 1)
      ans = add(ans, MOD - p2[dp[1][i]]);
    else
      ans = add(ans, p2[dp[1][i]]);
  }
  printf("%d\n", ans);
}