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 L = 20, N = 1e6 + 5;
const long long MOD = 1e9 + 7;
int dp[1 << L], n;
long long pow2[N];
int main() {
  scanf("%d", &n);
  int a;
  for (int i = (int)1; i <= n; i++) {
    scanf("%d", &a);
    dp[a]++;
  }
  pow2[0] = 1;
  for (int i = (int)1; i <= N - 1; i++) pow2[i] = pow2[i - 1] * 2 % MOD;
  for (int i = (int)0; i <= L - 1; i++)
    for (int msk = (int)0; msk <= (1 << L) - 1; msk++)
      if (msk >> i & 1 ^ 1) dp[msk] += dp[msk ^ (1 << i)];
  for (int msk = (int)0; msk <= (1 << L) - 1; msk++) {
  }
  long long ans = (pow2[dp[0]] - 1 + MOD) % MOD;
  for (int msk = (int)1; msk <= (1 << L) - 1; msk++) {
    if (dp[msk] == 0) continue;
    if (__builtin_popcount(msk) % 2) {
      ans = (ans - pow2[dp[msk]] + 1 + MOD) % MOD;
    } else {
      ans = (ans + pow2[dp[msk]] - 1 + MOD) % MOD;
    }
  }
  cout << ans << endl;
  return 0;
}