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;
int n, dp[1 << 20], num[1 << 20], mi[1 << 20];
long long ans;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    dp[x]++;
  }
  mi[0] = 1;
  for (int i = 1; i < 1 << 20; i++)
    num[i] = num[i >> 1] + (i & 1), mi[i] = mi[i - 1] * 2 % 1000000007;
  for (int j = 1; j <= 20; j++)
    for (int i = 0; i < 1 << 20; i++)
      if (i & (1 << j - 1)) dp[i ^ (1 << j - 1)] += dp[i];
  for (int i = 0; i < 1 << 20; i++) {
    if (num[i] & 1)
      ans = (ans - 1ll * mi[dp[i]] + 1) % 1000000007;
    else
      ans = (ans + 1ll * mi[dp[i]] - 1) % 1000000007;
  }
  printf("%lld", (ans + 1000000007) % 1000000007);
  return 0;
}