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 lint = long long;
const lint mod = 1e9 + 7;
#define all(x) (x).begin(), (x).end()
#define bitcount(n) __builtin_popcountl((lint)(n))
#define fcout cout << fixed << setprecision(15)
#define highest(x) (63 - __builtin_clzl(x))
template<class T> inline void YES(T condition){ if(condition) cout << "YES" << endl; else cout << "NO" << endl; }
template<class T> inline void Yes(T condition){ if(condition) cout << "Yes" << endl; else cout << "No" << endl; }
template<class T = string, class U = char>int character_count(T text, U character){ int ans = 0; for(U i: text){ ans += (i == character); } return ans; }
lint power(lint base, lint exponent, lint module){ if(exponent % 2){ return power(base, exponent - 1, module) * base % module; }else if(exponent){ lint root_ans = power(base, exponent / 2, module); return root_ans * root_ans % module; }else{ return 1; }}
struct position{ int y, x; }; position mv[4] = {{0, -1}, {1, 0}, {0, 1}, {-1, 0}}; // double euclidean(position first, position second){ return sqrt((second.x - first.x) * (second.x - first.x) + (second.y - first.y) * (second.y - first.y)); }
template<class T, class U> string to_string(pair<T, U> x){ return to_string(x.first) + "," + to_string(x.second); } string to_string(string x){ return x; }
template<class itr> void array_output(itr start, itr goal){ string ans; for(auto i = start; i != goal; i++) ans += to_string(*i) + " "; if(!ans.empty()) ans.pop_back(); cout << ans << endl; }
template<class itr> void cins(itr first, itr last){ for(auto i = first; i != last; i++){ cin >> (*i); } }
template<class T> T gcd(T a, T b){ if(a && b){ return gcd(min(a, b), max(a, b) % min(a, b)); }else{ return a; }} template<class T> T lcm(T a, T b){ return a / gcd(a, b) * b; }
struct combination{ vector<lint> fact, inv; combination(int sz) : fact(sz + 1), inv(sz + 1){ fact[0] = 1; for(int i = 1; i <= sz; i++){ fact[i] = fact[i - 1] * i % mod; } inv[sz] = power(fact[sz], mod - 2, mod); for(int i = sz - 1; i >= 0; i--){ inv[i] = inv[i + 1] * (i + 1) % mod; } } lint C(int p, int q) const{ if(q < 0 || p < q) return 0; return (fact[p] * inv[q] % mod * inv[p - q] % mod); } };
template<class itr> bool next_sequence(itr first, itr last, int max_bound){ itr now = last; while(now != first){ now--; (*now)++; if((*now) == max_bound){ (*now) = 0; }else{ return true; } } return false; }
template<class itr, class itr2> bool next_sequence2(itr first, itr last, itr2 first2, itr2 last2){ itr now = last; itr2 now2 = last2; while(now != first){ now--, now2--; (*now)++; if((*now) == (*now2)){ (*now) = 0; }else{ return true; } } return false; }

int dp[21][1 << 20];

int main(){
    int N;
    cin >> N;
    int cnt[1 << 20];
    fill(cnt, cnt + (1 << 20), 0);
    for(int i = 0; i < N; i++){
        int a;
        cin >> a;
        cnt[a]++;
    }
    for(int i = 0; i < (1 << 20); i++){
        dp[0][i] = cnt[i];
    }
    for(int i = 0; i < 20; i++){
        for(int j = 0; j < (1 << 20); j++){
            if((~(j >> i)) & 1){
                dp[i + 1][j] = dp[i][j] + dp[i][j | (1 << i)];
            }else{
                dp[i + 1][j] = dp[i][j];
            }
        }
    }
    //array_output(dp[20], dp[20] + 32);
    lint ans = 0;
    for(int i = 0; i < (1 << 20); i++){
        if(bitcount(i) % 2 == 0){
            ans += power(2, dp[20][i], mod);
        }else{
            ans -= power(2, dp[20][i], mod);
        }
        ans = (ans + mod) % mod;
    }
    cout << ans << endl;
}