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 mxlg = 21, MAXX = (1 << mxlg), MOD = 1e9 + 7;
int n, pw2[MAXX], dp[MAXX][mxlg], ans;
int sum(int x, int y) { return ((((x + y) % MOD) + MOD) % MOD); }
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n, pw2[0] = 1;
  for (int i = 1; i < MAXX; i++) pw2[i] = (pw2[i - 1] * 2) % MOD;
  for (int i = 0, x; i < n; i++) cin >> x, dp[x][0]++;
  for (int mask = 1; mask < mxlg; mask++) {
    for (int i = 0; i < MAXX; i++) {
      dp[i][mask] = dp[i][mask - 1];
      if (!(i & (1 << (mask - 1))))
        dp[i][mask] = (dp[i][mask] + dp[i + (1 << (mask - 1))][mask - 1]);
    }
  }
  for (int i = 0; i < MAXX; i++) {
    int x = __builtin_popcount(i), y = pow(-1, x);
    ans = sum(ans, (y * pw2[dp[i][mxlg - 1]]));
  }
  return cout << ans << endl, 0;
}