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;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
int ar[1048600];
int dp[1048600][21];
long long pow2[1048600];
int func(int x, int bit) {
  if (bit == -1) {
    return ar[x];
  }
  if (dp[x][bit] != -1) return dp[x][bit];
  int z = x & (1 << bit);
  int ans = 0;
  if (z) {
    ans = func(x, bit - 1);
  } else {
    ans = func(x, bit - 1) + func(x | (1 << bit), bit - 1);
  }
  return dp[x][bit] = ans;
}
int ff(int x) { return func(x, 19); }
void solve() {
  int n;
  cin >> n;
  memset(dp, -1, sizeof(dp));
  ;
  memset(ar, 0, sizeof(ar));
  ;
  for (int i = 0; i < (int)(n); i++) {
    int x;
    cin >> x;
    ar[x]++;
  }
  int mask = 1 << 20;
  long long ans = 0;
  pow2[0] = 1;
  for (int i = 1; i < 1000005; i++) {
    pow2[i] = (pow2[i - 1] * 2) % 1000000007;
  }
  while (mask > 0) {
    mask--;
    long long z = pow2[ff(mask)];
    int cnt = 0;
    int m = mask;
    while (m > 0) {
      if (m & 1) cnt = !cnt;
      m >>= 1;
    }
    if (cnt) z = (-z + 1000000007) % 1000000007;
    ans = (ans + z) % 1000000007;
  }
  cout << ans;
}
int main() {
  cin.sync_with_stdio(0);
  cout.sync_with_stdio(0);
  int TC = 1;
  for (int ZZ = 1; ZZ <= TC; ZZ++) {
    solve();
  }
  return 0;
}