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 MAXN = 1100000;
const int MOD = 1e9 + 7;
int val[MAXN];
int dp[21][(1 << 21)];
int GetLen(int x) {
  int rnt = 0;
  for (int i = 0; i < 20; i++)
    if ((1 << i) & x) rnt++;
  return rnt;
}
long long Pow(int x) {
  long long rnt = 1, mulv = 2;
  while (x) {
    if (x & 1) rnt *= mulv;
    rnt %= MOD;
    mulv *= mulv;
    mulv %= MOD;
    x >>= 1;
  }
  return rnt;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &val[i]);
    dp[0][val[i]]++;
  }
  for (register int i = 0; i < 20; ++i) {
    for (register int j = 0; j < (1 << 20); ++j) {
      if ((1 << i) & j)
        dp[i + 1][j] = dp[i][j];
      else
        dp[i + 1][j] = (dp[i][j] + dp[i][j + (1 << i)]) % MOD;
    }
  }
  long long ans = 0;
  for (register int i = 0; i < (1 << 20); ++i) {
    int len = GetLen(i);
    ans += Pow(dp[20][i]) * (len & 1 == 1 ? -1 : 1);
    ans = (ans % MOD + MOD) % MOD;
  }
  printf("%lld", ans);
  return 0;
}