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 long long INF = 1e18, MOD = 1e9 + 7, MOD2 = 1e6 + 3;
const int M = 1e6;
int cnt[1 << 20], cnt_r[1 << 20], n, d[1 << 20];
int bin[1000000];
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    d[a]++;
  }
  bin[0] = 1;
  for (int i = 1; i <= n; i++) {
    bin[i] = bin[i - 1] + bin[i - 1];
    if (bin[i] >= MOD) bin[i] -= MOD;
  }
  int m = 20;
  for (int i = 0; i < (1 << m); i++) {
    if (!d[i]) continue;
    for (int j = i; j & ((1 << 10) - 1); j = (i & (j - 1))) {
      cnt_r[j] += d[i];
    }
    cnt_r[(i >> 10) << 10] += d[i];
  }
  for (int i = 0; i < (1 << 20); i++) {
    if (!cnt_r[i]) continue;
    int hi = (i >> 10), lo = (i & ((1 << 10) - 1));
    for (int j = hi; j; j = (hi & (j - 1))) cnt[(j << 10) + lo] += cnt_r[i];
    cnt[lo] += cnt_r[i];
  }
  long long ans = bin[n] - 1;
  for (int i = 1; i < (1 << 20); i++) {
    if (!cnt[i]) continue;
    if (__builtin_popcount(i) & 1)
      ans += MOD - bin[cnt[i]] + 1;
    else
      ans += bin[cnt[i]] - 1;
    if (ans >= MOD) ans -= MOD;
  }
  cout << ans;
}