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 INF = 1LL << 60;
const double INFF = 1e100;
const double eps = 1e-8;
const int mod = 1000000007;
const int NN = 100005;
const int MM = 1000010;
int dp[20][1 << 20];
int a[1 << 20];
int q_pow(int x, int n, long long mod) {
  long long ans = 1, xx = x;
  for (; n > 0; n >>= 1) {
    if (n & 1) ans = ans * xx % mod;
    xx = xx * xx % mod;
  }
  return (int)ans;
}
int main() {
  int n, i, j, mask = 1 << 20;
  int ans, t, tt;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d", &a[i]);
  memset(dp, 0, sizeof(dp));
  for (i = 1; i <= n; i++) {
    if (a[i] & 1) {
      dp[0][a[i]]++;
      dp[0][a[i] ^ 1]++;
    } else {
      dp[0][a[i]]++;
    }
  }
  for (i = 0; i < 19; i++)
    for (j = 0; j < mask; j++) {
      t = 1 << (i + 1);
      if (j & t) {
        dp[i + 1][j] += dp[i][j];
        dp[i + 1][j - t] += dp[i][j];
      } else {
        dp[i + 1][j] += dp[i][j];
      }
    }
  ans = 0;
  for (j = 0; j < mask; j++) {
    t = 1;
    for (i = 0; i < 20; i++)
      if ((1 << i) & j) t = -t;
    tt = q_pow(2, dp[19][j], mod);
    tt--;
    if (tt < 0) tt += mod;
    ans += t * tt;
    if (ans >= mod) ans -= mod;
    if (ans < 0) ans += mod;
  }
  printf("%d\n", ans);
  return 0;
}