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 LN = 20;
const int MOD = 1000000007;
int n;
int arr[1 << LN];
int dp[1 << LN][LN + 1];
int f[1 << LN], p[1 << LN];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", arr + i);
    ++dp[arr[i]][0];
  }
  for (int mask = (1 << LN) - 1; mask >= 0; mask--) {
    for (int i = 0; i < LN; i++) {
      if (mask & (1 << i))
        dp[mask][i + 1] = dp[mask][i];
      else
        dp[mask][i + 1] = dp[mask][i] + dp[mask | (1 << i)][i];
    }
    f[mask] = dp[mask][LN];
  }
  p[0] = 1;
  for (int i = 1; i < (1 << LN); i++) {
    p[i] = (2LL * p[i - 1]) % MOD;
  }
  for (int i = 0; i < (1 << LN); i++) {
    p[i] = (p[i] - 1 + MOD) % MOD;
  }
  int ans = 0;
  for (int i = 0; i < (1 << LN); i++) {
    if (__builtin_popcount(i) & 1)
      ans = (ans - p[f[i]] + MOD) % MOD;
    else
      ans = (ans + p[f[i]]) % MOD;
  }
  printf("%d\n", ans);
}