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>
long long v[1000005];
long long dp[1000005];
long long f[25];
int n;
int main() {
  v[0] = 1;
  scanf("%d", &n);
  for (int i = 1; i <= 1000000; i++) {
    v[i] = (v[i - 1] << 1) % 1000000007;
  }
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    dp[x]++;
  }
  for (int j = 0; j <= 20; j++) {
    for (int i = 1; i <= 1000000; i++) {
      if (((1 << j) & i)) {
        dp[i ^ (1 << j)] += dp[i];
      }
    }
  }
  long long ans = 0;
  for (int i = 0; i <= 1000000; i++) {
    int cnt = 0;
    for (int j = 0; j <= 20; j++) {
      if ((1 << j) & i) {
        cnt++;
      }
    }
    f[cnt] += ((v[dp[i]] - 1) % 1000000007 + 1000000007) % 1000000007;
  }
  for (int i = 0; i <= 20; i++) {
    if (i % 2) {
      ans -= f[i];
      ans %= 1000000007;
    } else {
      ans += f[i];
      ans %= 1000000007;
    }
  }
  printf("%I64d\n", (ans % 1000000007 + 1000000007) % 1000000007);
  return 0;
}