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 maxK = 20;
const int maxN = (1 << maxK);
const int INF = (int)1e9;
const long long MOD = 7 + (long long)1e9;
const double eps = 1e-9;
int A[maxN];
int F[maxN];
int bits(int m) {
  int res = 0;
  for (int i = 0; i < maxK; i++) {
    res += (m >> i) & 1;
  }
  return res;
}
long long pow2[maxN + 5];
int main(void) {
  int n;
  cin >> n;
  pow2[0] = 1;
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    F[x]++;
    pow2[i + 1] = (pow2[i] * 2LL) % MOD;
  }
  for (int i = maxK - 1; i >= 0; i--) {
    for (int mask = (1 << i); mask < maxN; mask++) {
      if ((1 << i) & mask) {
        F[mask ^ (1 << i)] += F[mask];
      }
    }
  }
  long long res = 0;
  for (int mask = 0; mask < maxN; mask++) {
    if (bits(mask) & 1) {
      res = (res - (pow2[F[mask]] - 1) + MOD) % MOD;
    } else {
      res = (res + (pow2[F[mask]] - 1) + MOD) % MOD;
    }
  }
  cout << res << endl;
  return 0;
}