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 M = (1 << 20);
const long long MOD = 1e9 + 7;
int dp[M];
long long p2[M], ans;
int32_t main() {
  int n;
  scanf("%d", &n);
  p2[0] = 1;
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    dp[x]++;
    p2[i + 1] = (p2[i] * 2ll) % MOD;
  }
  for (int i = 20; ~i; i--)
    for (int mask = M; --mask;)
      if (mask & (1 << i)) dp[mask ^ (1 << i)] += dp[mask];
  for (int i = 0; i < M; i++)
    ans += p2[dp[i]] * ((int)__builtin_popcount(i) & 1 ? -1 : 1) + MOD,
        ans %= MOD;
  return cout << ans << endl, 0;
}