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 Set(int N, int pos) { return N = N | (1 << pos); }
int reset(int N, int pos) { return N = N & ~(1 << pos); }
bool check(int N, int pos) { return (bool)(N & (1 << pos)); }
long long po[1000000 + 5];
int dp[1000000 + 3][21], cnt[1000000 + 3];
int dp_func(int mask, int pos) {
  if (mask > 1000000) return 0;
  if (pos == -1) return cnt[mask];
  if (dp[mask][pos] != -1) return dp[mask][pos];
  int ret = 0;
  if (check(mask, pos) == 0) ret += dp_func(Set(mask, pos), pos - 1);
  ret += dp_func(mask, pos - 1);
  return dp[mask][pos] = ret;
}
int main() {
  int i, j, k, l, m, n;
  po[0] = 1;
  for (int i = 1; i <= 1000000; i++) {
    po[i] = po[i - 1] * 2;
    po[i] %= 1000000007;
  }
  memset(dp, -1, sizeof dp);
  scanf("%d", &n);
  int ma = 0;
  for (i = 1; i <= n; i++) {
    scanf("%d", &l);
    cnt[l]++;
    ma = max(ma, l);
  }
  long long ans = 0;
  for (int i = 0; i <= ma; i++) {
    int l = dp_func(i, 19);
    if (l == 0) continue;
    long long d = po[l] - 1;
    if (d < 0) d = (d + 1000000007) % 1000000007;
    if (__builtin_popcount(i) & 1)
      ans -= d;
    else
      ans += d;
    if (ans < 0) ans = (ans + 1000000007) % 1000000007;
    ans = ans % 1000000007;
  }
  cout << ans << endl;
}