#include<bits/stdc++.h>
using namespace std; 
char nl = '\n';
using i64 = long long;

void solve(int t) {
    // cout << "test #" << t << << nl;
    int n; cin >> n;
    vector<int> A(n);
    for(auto& a : A) cin >> a;

    // nxt[i][bit] stores the nearest position where this bit is set
    vector<vector<int>> nxt(n + 1, vector<int> (21, n));
    for(int i = n - 1; i >= 0; i--) {
        nxt[i] = nxt[i + 1];
        for(int b = 0; b < 21; b++) {
            if(A[i] & (1 << b)) nxt[i][b] = i;
        }
    }

    // prefix or can have atmost 20 distinct values
    // finding distinct elemnt in all prefix i..n
    vector<int> mark(int(1 << 21));
    for(int i = 0; i < n; i++) {
        // currently on prefix i..n
        int cur = A[i];
        int jump = i;
        // this while loop will run atmost 20 times
        while(jump < n) {
            cur |= A[jump];
            mark[cur] = 1;
            /*
                mn is the nearest position where we get new
                set bit which is currently unset in prefi i..jump
            */ 
            int mn = n;
            for(int b = 0; b < 21; b++) {
                if((cur & (1 << b)) == 0) mn = min(mn, nxt[jump][b]);
            }
            jump = mn;
        }
    }
    cout << accumulate(mark.begin(), mark.end(), 0) << nl;
} 

int main() {
    int tt = 1;
    // cin >> tt;
    for(int t = 1; t <= tt; t++) solve(t);
}