Jzzhu and Numbers

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 modulo1000000007 (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

input

4
0 1 2 3

output

input

6
5 2 0 5 2 1

output

53
分析：暴力肯定不行了，考虑容斥；
　　　答案为ans=Σ(-1)pow(2,num[x])，f(x)代表x中二进制1的个数，num[x]代表a[k]&x=x的个数；
　　　求num[x]=Σcnt[k](a[k]&x=x)，即为高维前缀和，与spoj tle类似；
代码：

#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <cmath>

#include <algorithm>

#include <climits>

#include <cstring>

#include <string>

#include <set>

#include <bitset>

#include <map>

#include <queue>

#include <stack>

#include <vector>

#define rep(i,m,n) for(i=m;i<=n;i++)

#define mod 1000000007

#define inf 0x3f3f3f3f

#define vi vector<int>

#define pb push_back

#define mp make_pair

#define fi first

#define se second

#define ll long long

#define pi acos(-1.0)

#define pii pair<int,int>

#define sys system("pause")

const int maxn=1e6+;

const int N=2e2+;

using namespace std;

ll gcd(ll p,ll q){return q==?p:gcd(q,p%q);}

ll qpow(ll p,ll q){ll f=;while(q){if(q&)f=f*p;p=p*p;q>>=;}return f;}

int n,m,k,t,num[<<];

ll dp[<<],p[maxn];

int main()

{

    int i,j;

    p[]=;

    rep(i,,maxn-)p[i]=p[i-]*%mod;

    scanf("%d",&n);

    rep(i,,n)scanf("%d",&j),dp[j]++;

    rep(i,,)

    {

        rep(j,,(<<)-)

        {

            if((~j)&(<<i))(dp[j]+=dp[j^(<<i)])%=mod;

            if(j&(<<i))num[j]++;

        }

    }

    ll ret=;

    rep(i,,(<<)-)

    {

        if(num[i]&)ret=(ret-p[dp[i]]+mod)%mod;

        else ret=(ret+p[dp[i]])%mod;

    }

    printf("%lld\n",ret);

    return ;

}