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 = 21;
const int M = 1e6 + 1;
int mod = 1e9 + 7;
int m[N][M], B[M], C[N];
int main() {
  int n;
  int a;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a), m[0][a]++;
  B[0] = 1;
  for (int i = 1; i <= n; i++) B[i] = 1ll * 2 * B[i - 1] % mod;
  for (int i = 0; i <= n; i++) B[i] = (mod + B[i] - 1) % mod;
  for (int i = M - 1; i > -1; i--) {
    int cur = 0;
    for (int j = 1; j < N; j++) m[j][i] /= j, cur += m[j][i];
    for (int k = 0; k < N; k++) {
      if (((i >> k) & 1))
        for (int j = 0; j < N - 1; j++) m[j + 1][i - (1 << k)] += m[j][i];
    }
    m[0][i] += cur;
  }
  int ans = 0;
  for (int i = 0; i < M; i++) {
    int cur = m[0][i];
    if (__builtin_popcount(i) % 2 == 1)
      ans = (ans + mod - B[cur]) % mod;
    else
      ans = (ans + B[cur]) % mod;
  }
  printf("%d", ans);
}