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 = 1e6 + 10, M = 21, MX = 1 << (M - 1), NX = 1e9 + 7;
int n, r, t, c, A[N], P[N], dp[M][1 << M];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &A[i]), dp[0][A[i]]++;
  P[0] = 1;
  for (int i = 1; i <= n; i++) P[i] = P[i - 1] * 2 % NX;
  for (int i = 1; i < M; i++)
    for (int j = 0; j < MX; j++) {
      dp[i][j] = dp[i - 1][j];
      if (!(j & (1 << (i - 1)))) dp[i][j] += dp[i - 1][j + (1 << (i - 1))];
    }
  for (int j = 0; j < MX; j++) {
    if (__builtin_popcount(j) & 1)
      c = (c - P[dp[M - 1][j]] + NX) % NX;
    else
      c = (c + P[dp[M - 1][j]]) % NX;
  }
  return !printf("%d", c);
}