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 up = (1 << 20);
const int MOD = (int)1e9 + 7;
int dp[2][up];
int modExp(int a, int b) {
  long long t = 1, y = a;
  while (b) {
    if (b & 1) t = t * y % MOD;
    y = y * y % MOD;
    b >>= 1;
  }
  return (int)t;
}
int main() {
  int n, t, ans = 0;
  scanf("%d", &n);
  int pre = 0, now = 1;
  for (int i = 0; i < n; i++) {
    scanf("%d", &t);
    dp[pre][t]++;
  }
  for (int i = 0; i < 20; i++) {
    memset(dp[now], 0, sizeof(dp[now]));
    for (int j = 0; j < up; j++) {
      if (j >> i & 1) {
        dp[now][j] = dp[pre][j];
      } else {
        dp[now][j] = dp[pre][j] + dp[pre][j + (1 << i)];
        dp[now][j] %= MOD;
      }
    }
    pre ^= 1, now ^= 1;
  }
  for (int i = 0; i < up; i++) {
    int cnt = __builtin_popcount(i);
    int res = modExp(2, dp[pre][i]);
    ans += (cnt & 1) ? -res : res;
    ans = (ans % MOD + MOD) % MOD;
  }
  printf("%d\n", ans);
  return 0;
}