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;
int sum[1 << 20];
const int P = 1e9 + 7;
int power(int a, int b) {
  int ans = 1, z = a;
  for (; b; b >>= 1, z = (long long)z * z % P)
    if (b & 1) ans = (long long)ans * z % P;
  return ans;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    int x;
    scanf("%d", &x);
    ++sum[x];
  }
  for (int i = 0; i < 20; ++i)
    for (int j = 0; j < 1 << 20; ++j)
      if (j >> i & 1) sum[j ^ (1 << i)] += sum[j];
  int ans = 0;
  for (int opt = 0; opt < 1 << 20; ++opt) {
    int ans1 = power(2, sum[opt]) - 1;
    if (ans1 < 0) ans1 += P;
    int cur = 0;
    for (int i = opt; i; i -= i & -i) cur ^= 1;
    if (!cur)
      ans += ans1;
    else
      ans -= ans1;
    if (ans < 0) ans += P;
    ans %= P;
  }
  printf("%d\n", ans);
  return 0;
}