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 mod = 1e9 + 7;
const int maxn = 1e6 + 5;
int n;
int a[maxn];
int dp[(1 << 21)][21];
long long powmod(long long x, long long y, long long m) {
  long long z = x % m;
  if (!y) return 1;
  if (y & 1)
    return z * powmod(z * z, (y - 1) / 2, m) % m;
  else
    return powmod(z * z, y / 2, m) % m;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < n; i++) dp[a[i]][0]++;
  for (int i = 1; i < 21; i++) {
    for (int j = 0; j < (1 << 21); j++) {
      if ((j >> (i - 1)) & 1)
        dp[j][i] = dp[j][i - 1];
      else
        dp[j][i] = dp[j][i - 1] + dp[j + (1 << (i - 1))][i - 1];
    }
  }
  long long ans = 0;
  for (int i = 0; i < (1 << 21); i++) {
    long long cur = powmod(2, dp[i][20], mod) - 1;
    if (__builtin_popcount(i) & 1) cur = -cur;
    ans += cur + mod;
    ans %= mod;
  }
  cout << ans << endl;
  return 0;
}