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 N = 1 << 20;
const long long MOD = 1e9 + 7;
int n, m;
long long dp[21][N];
long long g[N];
bool s[N];
void init() {
  scanf("%d", &n);
  g[0] = 1;
  for (int i = 1; i <= n; i++) g[i] = (g[i - 1] << 1LL) % MOD;
  int f0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &f0);
    dp[0][f0]++;
  }
}
void work() {
  for (int i = 1; i <= 20; i++) {
    for (int j = 0; j < N; j++)
      if ((j & (1 << (i - 1))) == 0)
        dp[i][j] = dp[i - 1][j] + dp[i - 1][j | (1 << (i - 1))];
      else
        dp[i][j] = dp[i - 1][j], s[j] ^= 1;
  }
  long long ans = 0;
  for (int i = 0; i < N; i++)
    ans = (ans + (s[i] ? (-1LL) : 1LL) * g[dp[20][i]] + MOD) % MOD;
  printf("%I64d\n", ans);
}
int main() {
  init();
  work();
  return 0;
}