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 dp[21][1 << 20];
long long mod_pow(long long a, int b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % 1000000007;
    a = a * a % 1000000007;
    b >>= 1;
  }
  return ret;
}
int main() {
  int n, x;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &x);
    ++dp[0][x];
  }
  for (int i = 0; i <= 19; ++i) {
    for (int mask = 0; mask < (1 << 20); ++mask) {
      dp[i + 1][mask] += dp[i][mask];
      if (dp[i + 1][mask] >= 1000000007) dp[i + 1][mask] -= 1000000007;
      if (mask & (1 << i)) {
        dp[i + 1][mask ^ (1 << i)] += dp[i][mask];
        if (dp[i + 1][mask ^ (1 << i)] >= 1000000007)
          dp[i + 1][mask ^ (1 << i)] -= 1000000007;
      }
    }
  }
  int ans = 0;
  for (int mask = 0; mask < (1 << 20); ++mask) {
    if (__builtin_popcount(mask) % 2 == 0) {
      ans += mod_pow(2, dp[20][mask]);
      if (ans >= 1000000007) ans -= 1000000007;
    } else {
      ans -= mod_pow(2, dp[20][mask]);
      if (ans < 0) ans += 1000000007;
    }
  }
  printf("%d\n", ans);
  return 0;
}