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 long long maxn = 1e6 + 3, mod = 1e9 + 7;
long long n, ans = 0;
long long v[maxn];
long long dp[(1 << 20) + 1];
long long binexp(long long b, long long e) {
  long long x = 1;
  long long y = b;
  while (e) {
    if (e & 1) x = (x * y) % mod;
    y = (y * y) % mod;
    e >>= 1;
  }
  return x;
}
int main() {
  scanf("%lld", &n);
  for (long long i = 1; i <= n; i++) {
    scanf("%lld", &v[i]);
    dp[(1 << 20) - v[i] - 1]++;
  }
  for (long long i = 0; i <= 20; i++)
    for (long long j = 0; j < (1 << 20); j++)
      if (j & (1 << i)) {
        dp[j] += dp[j ^ (1 << i)];
        if (dp[j] >= mod) dp[j] -= mod;
      }
  for (long long i = 0; i < (1 << 20); i++) {
    long long bits = 20 - __builtin_popcount(i);
    if (bits & 1) {
      ans -= binexp(2, dp[i]);
      if (ans < 0) ans += mod;
    } else {
      ans += binexp(2, dp[i]);
      if (ans >= mod) ans -= mod;
    }
  }
  printf("%lld", ans);
  return 0;
}