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 M = 1048576;
int f[M + 10], pw[M + 10];
int siz[M + 10];
int main() {
  int n;
  scanf("%d", &n);
  pw[0] = 1;
  for (int i = 1; i <= n; i++) {
    int a;
    scanf("%d", &a);
    f[a]++;
    pw[i] = pw[i - 1] << 1;
    if (pw[i] >= 1000000007) pw[i] -= 1000000007;
  }
  for (int step = 1; step < M; step <<= 1)
    for (int i = 0; i < M; i += step << 1)
      for (int j = i; j < i + step; j++) {
        f[j] += f[j + step];
        if (f[j] >= 1000000007) f[j] -= 1000000007;
      }
  for (int i = 1; i < M; i++) siz[i] = siz[i >> 1] + (i & 1);
  int ans = 0;
  for (int i = 0; i < M; i++) {
    if (siz[i] & 1)
      ans -= pw[f[i]];
    else
      ans += pw[f[i]];
    if (ans >= 1000000007) ans -= 1000000007;
    if (ans < 0) ans += 1000000007;
  }
  printf("%d\n", ans);
  return 0;
}