Problem: Zero AND Subsets

Given a multiset of nonnegative integers a_1, a_2, .., a_N.

How many non-empty subsets of this set have a value bitwiseAND of 0?

Find the remainder of the answer divided by 10 ^ 9 + 7.

input


N
a_1 a_2 ... a_N


output

Divide the answer by 10 ^ 9 + 7 and output the remainder.

Constraint

* 1 \ leq N \ leq 10 ^ 5
* 0 \ leq a_i \ leq 2 ^ {20} -1



Input example


6
8 6 9 1 2 1


Output example


51






Example

Input

6
8 6 9 1 2 1


Output

51

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
template<class T,class U> using P = pair<T,U>;
template<class T> using vec = vector<T>;
template<class T> using vvec = vector<vec<T>>;

constexpr ll mod = 1e9+7;
struct mint {
    ll x;
    mint(ll x=0):x((x%mod+mod)%mod){}
    mint& operator+=(const mint a) {
        if ((x += a.x) >= mod) x -= mod;
        return *this;
    }
    mint& operator-=(const mint a) {
        if ((x += mod-a.x) >= mod) x -= mod;
        return *this;
    }
    mint& operator*=(const mint a) {
        (x *= a.x) %= mod;
        return *this;
    }
    mint operator+(const mint a) const {
        mint res(*this);
        return res+=a;
    }
    mint operator-(const mint a) const {
        mint res(*this);
        return res-=a;
    }
    mint operator*(const mint a) const {
        mint res(*this);
        return res*=a;
    }
    mint pow(ll t) const {
        if (!t) return 1;
        mint a = pow(t>>1);
        a *= a;
        if (t&1) a *= *this;
        return a;
    }
    // for prime mod
    mint inv() const {
        return pow(mod-2);
    }
    mint& operator/=(const mint a) {
        return (*this) *= a.inv();
    }
    mint operator/(const mint a) const {
        mint res(*this);
        return res/=a;
    }
};

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int N;
    cin >> N;
    vec<int> g(1<<20,0);
    for(int i=0;i<N;i++){
        int a;
        cin >> a;
        g[a]++;
    }
    for(int i=0;i<20;i++){
        for(int S=(1<<20)-1;S>=0;S--) if(!(S&(1<<i))) g[S] += g[S^(1<<i)];
    }
    mint two = 2;
    mint ans = 0;
    for(int S=0;S<(1<<20);S++){
        int d = __builtin_popcount(S);
        if(d%2==0) ans += two.pow(g[S])-1;
        else ans -= two.pow(g[S])-1;
    }
    cout << ans.x << endl;
}