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>
int FASTBUFFER;
using namespace std;
const int N = 1048576;
const int mo = 1000000007;
int dp[N * 2 + 5];
int mi[N + 5], fac[N + 5];
int n, ans;
int main() {
  scanf("%d", &n);
  mi[0] = 1;
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    dp[x]++;
    mi[i] = (mi[i - 1] + mi[i - 1]) % mo;
  }
  for (int i = 1; i < N; i <<= 1) {
    for (int j = 0; j < N; j++) {
      if (i & j) {
        dp[i ^ j] += dp[j];
      }
    }
  }
  ans = mi[n];
  fac[0] = 1;
  for (int i = 1; i < N; i++) {
    fac[i] = -fac[i - (i & (0 - i))];
    (ans += fac[i] * mi[dp[i]]) %= mo;
  }
  while (ans < 0) {
    ans += mo;
  }
  printf("%d\n", ans);
  return 0;
}