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 int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3fll;
const int MAXN = 2e6 + 10;
const int MOD = 1e9 + 7;
int cnt[MAXN], q[MAXN];
long long pow2[MAXN];
void build(int l, int r) {
  if (r - l == 1) {
    q[l] = cnt[l];
    return;
  }
  int mid = (l + r) / 2;
  build(l, mid);
  build(mid, r);
  for (int a = l; a < mid; ++a) q[a] = q[a] + q[a + mid - l];
}
int main() {
  int n;
  scanf("%d", &n);
  for (int a = 0; a < n; ++a) {
    int x;
    scanf("%d", &x);
    cnt[x]++;
  }
  pow2[0] = 1;
  for (int a = 1; a < MAXN; ++a) pow2[a] = (2 * pow2[a - 1]) % MOD;
  build(0, 1 << 20);
  long long ans = 0;
  for (int m = 0; m < (1 << 20); ++m) {
    if (__builtin_popcount(m) & 1)
      ans -= pow2[q[m]];
    else
      ans += pow2[q[m]];
    ans %= MOD;
  }
  cout << (ans + MOD) % MOD << endl;
  return 0;
}