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 = 1000000007;
const double eps = 1e-8;
const int nm = 1100005;
int n, k, m, t;
long long res;
int cnt[nm], f[nm][23];
int POW(int x, int z) {
  if (z == 0) return 1;
  if (z == 1) return x % MOD;
  int u = POW(x, z >> 1);
  u = ((long long)u * u) % MOD;
  if (z % 2) u = ((long long)u * x) % MOD;
  return u;
}
int main() {
  int i, j, x, y, z, w;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &x);
    cnt[x]++;
  }
  for (i = (1 << 20) - 1; i >= 0; i--) {
    if (i & 1)
      f[i][0] = cnt[i];
    else
      f[i][0] = cnt[i] + cnt[i + 1];
    for (j = 1; j < 20; j++) {
      if (i & (1 << j))
        f[i][j] = f[i][j - 1];
      else
        f[i][j] = f[i][j - 1] + f[i + (1 << j)][j - 1];
    }
  }
  for (i = (1 << 20) - 1; i >= 0; i--) {
    x = __builtin_popcount(i);
    if (x & 1)
      res -= POW(2, f[i][19]);
    else
      res += POW(2, f[i][19]);
    res = (res + MOD) % MOD;
  }
  cout << res;
  return 0;
}