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 N = 20;
const int V = (1 << 20) + 10;
const int P = 1000000007;
int Pow(int x, int y) {
  int ret = 1;
  while (y) {
    if (y & 1) ret = (long long)ret * x % P;
    x = (long long)x * x % P;
    y /= 2;
  }
  return ret;
}
void utf(int *a, int l, int r) {
  if (l + 1 != r) {
    int mid = (l + r) / 2;
    for (int i = 0; i < mid - l; i++) {
      a[l + i] -= a[mid + i];
      if (a[l + i] < 0) a[l + i] += P;
    }
    utf(a, l, mid);
    utf(a, mid, r);
  }
}
void tf(int *a, int l, int r) {
  if (l + 1 != r) {
    int mid = (l + r) / 2;
    tf(a, l, mid);
    tf(a, mid, r);
    for (int i = 0; i < mid - l; i++) {
      a[l + i] += a[mid + i];
      if (a[l + i] >= P) a[i + l] -= P;
    }
  }
}
int cnt[V], q[V];
int n, x, pro[V];
int pt[V];
int main() {
  pt[0] = 1;
  for (int i = 1; i < (1 << N); i++) pt[i] = pt[i - 1] * 2 % P;
  while (~scanf("%d", &n)) {
    memset(cnt, 0, sizeof(cnt));
    for (int i = 0; i < n; i++) {
      scanf("%d", &x);
      cnt[x]++;
    }
    for (int i = 0; i < N; i++) {
      for (int j = 0; j < (1 << N); j++) {
        if (j & (1 << i)) {
          cnt[j ^ (1 << i)] += cnt[j];
          if (cnt[j ^ (1 << i)] >= P) {
            cnt[j ^ (1 << i)] -= P;
          }
        }
      }
    }
    for (int i = 0; i < (1 << N); i++) pro[i] = pt[cnt[i]];
    for (int i = N - 1; i >= 0; i--) {
      for (int j = (1 << N); j >= 0; j--) {
        if (j & (1 << i)) {
          pro[j ^ (1 << i)] += P - pro[j];
          if (pro[j ^ (1 << i)] >= P) {
            pro[j ^ (1 << i)] -= P;
          }
        }
      }
    }
    pro[0] = (pro[0] % P + P) % P;
    printf("%d\n", pro[0]);
  }
  return 0;
}