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 maxn = 1200 * 1000 + 10, mod = 1000000007, maxk = 21;
int tvn[maxn], n, a, t[maxn], dp[maxn][maxk];
long long ans;
bool b[maxk];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a);
    t[a]++;
  }
  for (int i = 1; i < maxn; i++) tvn[i] = (((2 * tvn[i - 1]) % mod) + 1) % mod;
  for (int i = (1 << (maxk - 1)) - 1; i >= 1; i--) {
    a = i;
    for (int j = maxk - 1; j >= 1; j--) {
      b[j] = a & 1;
      a >>= 1;
    }
    dp[i][0] = t[i];
    for (int j = 1; j < maxk; j++) {
      dp[i][j] = dp[i][j - 1];
      if (!b[j]) dp[i][j] += dp[i ^ (1 << (maxk - j - 1))][j - 1];
    }
    if (__builtin_popcount(i) & 1)
      ans = (ans + tvn[dp[i][maxk - 1]]) % mod;
    else
      ans = (ans + mod - tvn[dp[i][maxk - 1]]) % mod;
  }
  printf("%I64d", (tvn[n] + mod - ans) % mod);
  return 0;
}