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>
template <typename Tp>
Tp Max(const Tp &a, const Tp &b) {
  return a > b ? a : b;
}
template <typename Tp>
Tp Min(const Tp &a, const Tp &b) {
  return a < b ? a : b;
}
template <typename Tp>
void Read(Tp &x) {
  Tp in = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    in = in * 10 + ch - '0';
    ch = getchar();
  }
  x = in * f;
}
const int SN = 2000000 + 10;
const int MOD = 1000000007;
int f[SN], a[SN], n;
int power[SN];
int main() {
  power[0] = 1;
  for (int i = 1; i <= SN - 10; i++) power[i] = (power[i - 1] * 2) % MOD;
  while (scanf("%d", &n) != EOF) {
    memset(f, 0, sizeof f);
    for (int i = 1; i <= n; i++) Read(a[i]), f[a[i]]++;
    for (int i = 1; i <= 20; i++)
      for (int sta = SN - 10; sta >= 0; sta--) {
        if ((sta >> (i - 1)) & 1) f[sta ^ (1 << (i - 1))] += f[sta];
      }
    long long ans = (power[n] - 1 + MOD) % MOD;
    for (int sta = 1; sta <= SN - 10; sta++) {
      int cnt = 0;
      for (int i = 1; i <= 20; i++)
        if ((sta >> (i - 1)) & 1) cnt++;
      if (cnt & 1)
        ans = (ans - (power[f[sta]] - 1 + MOD) % MOD + MOD) % MOD;
      else
        ans = (ans + (power[f[sta]] - 1) + MOD) % MOD;
    }
    printf("%I64d\n", ans);
  }
  return 0;
}