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 mod = 1e9 + 7;
int n, m, dp[2][1200005], f[1200005];
inline int exp(int x, int y) {
  int ans = 1;
  while (y) {
    if (y & 1) ans = (long long)ans * x % mod;
    x = (long long)x * x % mod;
    y >>= 1;
  }
  return ans;
}
int main() {
  scanf("%d", &n);
  int k = 0;
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    dp[k][x]++;
    m = max(m, x);
    f[x]++;
  }
  n = 2;
  while (n <= m) n <<= 1;
  for (int i = 1; i < 20; i++) {
    k ^= 1;
    for (int j = 0; j < n; j++) {
      dp[k][j] = 0;
      for (int l = 0; l < 20; l++)
        if (!(j & (1 << l))) dp[k][j] += dp[k ^ 1][j ^ (1 << l)];
      dp[k][j] /= i;
      f[j] = f[j] + dp[k][j];
    }
  }
  int ans = 0;
  for (int i = 0; i < n; i++) {
    if (f[i] < 0) return 0;
    int num = 0, t = exp(2, f[i]);
    for (int j = 0; j < 20; j++)
      if (i & (1 << j)) num++;
    if (num & 1)
      ans = (ans - t + mod) % mod;
    else
      ans = (ans + t) % mod;
  }
  printf("%d\n", ans);
}