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 N = 2020;
const int INF = 1000000000;
const long long Mod = 1000000007LL;
const int tot = (1 << 20);
int cnt[tot + 10];
void add(long long &x, long long y) {
  x += y;
  if (x >= Mod) x -= Mod;
}
long long pow_mod(long long a, long long k) {
  long long ret = 1;
  while (k) {
    if (k & 1) ret = ret * a % Mod;
    a = a * a % Mod;
    k >>= 1;
  }
  return ret;
}
int main() {
  int n;
  scanf("%d", &n);
  int cur = tot - 1;
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    cnt[x]++;
    cur = cur & x;
  }
  if (cur) {
    printf("0\n");
    return 0;
  }
  for (int i = 0; i < 20; i++) {
    for (int j = 0; j < tot; j++) {
      if (j & (1 << i)) {
        cnt[j ^ (1 << i)] += cnt[j];
      }
    }
  }
  long long ans = 0;
  for (int i = 0; i < tot; i++) {
    long long tmp = pow_mod(2LL, cnt[i]);
    if (__builtin_popcount(i) & 1) {
      ans = (ans - tmp + Mod) % Mod;
    } else {
      ans = (ans + tmp) % Mod;
    }
  }
  ans = (ans % Mod + Mod) % Mod;
  cout << ans << endl;
  return 0;
}