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 maxn = 1048576 + 10, M = 19, MOD = 1000000007;
int n, ans, f[maxn];
int Pow(long long a, int b) {
  long long res = 1;
  for (; b; b >>= 1, a = a * a % MOD)
    if (b & 1) res = (res * a) % MOD;
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    f[x]++;
  }
  for (int i = 0; i <= M; i++)
    for (int j = 0; j <= (1 << M + 1) - 1; j++)
      if (!((j >> i) & 1)) (f[j] += f[j | (1 << i)]) %= MOD;
  for (int i = 0; i <= (1 << M + 1) - 1; i++) {
    int cnt = 0;
    for (int j = 0; j <= M; j++)
      if ((i >> j) & 1) cnt++;
    (ans += ((cnt & 1) ? -1 : 1) * (Pow(2, f[i]) - 1)) %= MOD;
  }
  printf("%d\n", (ans + MOD) % MOD);
  return 0;
}