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;
void read(int &x) {
  x = 0;
  char c = getchar();
  for (; c < 48; c = getchar())
    ;
  for (; c > 47; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
}
const int M = 1 << 20;
long long po[M];
int dp[2][M], cnt[M], sz = sizeof(dp[0]), op[M];
int main() {
  int n;
  read(n);
  int i, x, now = 0;
  po[0] = 1;
  for (i = 1; i <= n; i++) po[i] = (po[i - 1] * 2) % 1000000007;
  long long res = 0;
  for (; n--;) {
    read(x);
    cnt[x]++;
    if (x & 1) {
      dp[now][x]++;
      dp[now][x ^ 1]++;
    } else {
      dp[now][x]++;
    }
  }
  int tmp, j;
  for (i = 1; i < 20; i++) {
    tmp = 1 << i;
    now = !now;
    memset(dp[now], 0, sz);
    for (j = 0; j < M; j++) {
      (dp[now][j] += dp[!now][j]) %= 1000000007;
      if (j & tmp) (dp[now][j ^ tmp] += dp[!now][j]) %= 1000000007;
    }
  }
  op[0] = 1;
  for (i = 0; i < M; i++) {
    if (i) op[i] = -op[i ^ (i & (-i))];
    res = (res + op[i] * (po[dp[now][i]] - 1)) % 1000000007;
  }
  printf("%lld\n", (res + 1000000007) % 1000000007);
  return 0;
}