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<iostream>
#include<algorithm>
#include<vector>
#include<string>
#include<set>
#include<queue>
#include<stack>
#include<bitset>
#include<functional>
#include<map>
#include<iomanip>
#include<limits>
#include<unordered_set> 
#include<cmath>
using namespace std;
//long long p = 998244353;
long long p = 1000000007;
#define int long long
#define vel vector<long long>
#define vvel vector<vel>
#define rep(i,n) for(int i=0;i<n;i++)
#define sor(v) sort(v.begin(),v.end())
#define mmax(a,b) a=max(a,b)
#define mmin(a,b) a=min(a,b)
#define mkp make_pair
#define pin pair<int,int>
#define qin pair<pin,int>
#define V vector
#define Endl endl
#define veb vector<bool>
#define fcout cout << fixed << setprecision(15)
#define rev(s) reverse(s.begin(),s.end())
#define lower(h,val) lower_bound(h.begin(),h.end(),val)-h.begin()
#define upper(h,val) upper_bound(h.begin(),h.end(),val)-h.begin()
int max_kai = 150000;
vel kai(max_kai, 1);
vel inv_kai;
int rui(int a, int n, int mod) {
    if (n == 0) { return 1 % mod; }
    int x = rui(a, n / 2, mod);
    x *= x; x %= mod;
    if (n % 2 == 1) { x *= a; x %= mod; }
    return x;
}
vel pa;
int root(int x) {
    if (pa[x] == -1) { return x; }
    int ans = root(pa[x]); pa[x] = ans;
    return ans;
}
void marge(int x, int y) {
    x = root(x);
    y = root(y);
    if (x != y) { pa[x] = y; }
}
int gcd(int x, int y) {
    if (x < y) { return gcd(y, x); }
    if (y == 0) { return x; }
    return gcd(y, x % y);
}
long long modinv(long long a, long long m) {
    long long b = m, u = 1, v = 0;
    while (b) {
        long long t = a / b;
        a -= t * b; swap(a, b);
        u -= t * v; swap(u, v);
    }
    u %= m;
    if (u < 0) u += m;
    return u;
}
vel uni(vel x) {
    if (x.size() == 0) { return x; }
    sor(x);
    int n = x.size();
    vel ans(1, x[0]);
    for (int j = 1; j < n; j++) {
        if (x[j - 1] != x[j]) { ans.push_back(x[j]); }
    }
    x = ans;
    return x;
}
void pr(vel& v) {
    int n = v.size();
    if (n != 0) {
        cout << v[0];
        rep(i, n - 1) {
            cout << " " << v[i + 1];
        }
        cout << endl;
    }
}
int sol(int x, int k) {
    if (x == 0 || k == 0) { return 0; }
    return x + sol(x / 2, k - 1);
}
vel pri(int n) {
    veb ans(n, true);
    vel prime;
    for (int j = 2; j < n; j++) {
        if (ans[j]) {
            prime.push_back(j);
            for (int k = 2; j * k < n; k++) {
                ans[j * k] = false;
            }
        }
    }
    return prime;
}
int solve(int L, int R, vel& prime) {//[L,R)
    if (L == 0) { L++; }
    int diff = R - L;
    vel num(diff);
    rep(i, diff) { num[i] = L + i; }
    veb ans(diff, true);
    for (auto p : prime) {
        int qL = (L + p - 1) / p;
        qL *= p;
        for (int i = qL; i < R; i += p) {
            int qi = i - L;
            while (num[qi] % p == 0) {
                ans[qi] = !ans[qi];
                num[qi] /= p;
            }
        }
    }
    int ret = 0;
    rep(i, diff) {
        if (num[i] != 1) { ans[i] = !ans[i]; }
        if (ans[i]) { ret++; }
        else { ret--; }
    }
    return ret;
}
vel dp;
int m;
vel rui2;
vel a;
void solve(int mod, int st,int to) {
    if (mod == 0) { return; }
    solve(mod/2, st,to);
    int mid1 = st + mod;
    int mid2 = mid1 + mod;
    int l = lower_bound(a.begin(), a.end(), mid1) - a.begin();
    int r = lower_bound(a.begin(), a.end(), mid2) - a.begin();
    dp[mid1] = rui2[r-l];
    solve(mod/2,mid1,mid2);
}
signed main() {
    int n; cin >> n;
    a=vel(n);
    rep(i, n) { cin >> a[i]; }
    m = (1 << 20);
    dp=vel(m,0);
    rep(i, n) { dp[a[i]]++; }
    rep(j, 20) {
        rep(i, m) {
            if ((i & (1 << j)) != 0) {
                int back = i - (1 << j);
                dp[back] += dp[i];
            }
        }
    }
    rui2 = vel(n + 1, 1);
    rep(i, n) {
        rui2[i + 1] = rui2[i] * 2;
        rui2[i + 1] %= p;
    }
    rep(i, n + 1) { rui2[i]--; }
    int ans = 0;
    for(int i=1;i<m;i++) {
        int sum = 0;
        rep(j, 20) {
            if ((i & (1 << j)) != 0) { sum++; }
        }
        if (sum % 2 == 1) { ans += rui2[dp[i]]; }
        else { ans += p - rui2[dp[i]]; }
        ans %= p;
    }
    ans = rui2[n] + p - ans; ans %= p;
    cout << ans << endl;
    return 0;
}