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 mod = 1000000007ll;
long long dp[(1 << 20)][21], p[(1 << 20)], a[(1 << 20)] = {0};
inline long long add(long long p, long long q) {
  long long ret = p + q;
  return ((ret >= mod) ? ret - mod : ret);
}
int main() {
  int n, l, lim, mx = 0;
  scanf("%d", &n);
  p[0] = 1;
  for (int i = 1; i < (1 << 20); ++i) p[i] = add(p[i - 1], p[i - 1]);
  for (int i = 0; i < n; ++i) {
    int b;
    scanf("%d", &b);
    mx = (((mx) > (b)) ? (mx) : (b));
    ++a[b];
  }
  l = log2(mx);
  lim = (1 << (l + 1));
  for (int i = 0; i < lim; ++i) {
    if (i & 1)
      dp[i][0] = a[i];
    else
      dp[i][0] = a[i] + a[i ^ 1];
  }
  for (int i = 1; i <= l; ++i) {
    for (int j = 0; j < lim; ++j) {
      if (j & p[i])
        dp[j][i] = dp[j][i - 1];
      else
        dp[j][i] = add(dp[j][i - 1], dp[j ^ p[i]][i - 1]);
    }
  }
  long long ans = 0, prv = 0;
  for (int i = 0; i < lim; ++i) {
    if (__builtin_popcount(i) & 1)
      ans = add(ans, mod - p[dp[i][l]]);
    else
      ans = add(ans, p[dp[i][l]]);
  }
  printf("%lld\n", ans);
  return 0;
}