<p>Given an integer array <code>arr</code>, return <em>the number of distinct bitwise ORs of all the non-empty subarrays of</em> <code>arr</code>.</p>

<p>The bitwise OR of a subarray is the bitwise OR of each integer in the subarray. The bitwise OR of a subarray of one integer is that integer.</p>

<p>A <strong>subarray</strong> is a contiguous non-empty sequence of elements within an array.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> arr = [0]
<strong>Output:</strong> 1
<strong>Explanation:</strong> There is only one possible result: 0.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> arr = [1,1,2]
<strong>Output:</strong> 3
<strong>Explanation:</strong> The possible subarrays are [1], [1], [2], [1, 1], [1, 2], [1, 1, 2].
These yield the results 1, 1, 2, 1, 3, 3.
There are 3 unique values, so the answer is 3.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> arr = [1,2,4]
<strong>Output:</strong> 6
<strong>Explanation:</strong> The possible results are 1, 2, 3, 4, 6, and 7.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= arr.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>0 &lt;= arr[i] &lt;= 10<sup>9</sup></code></li>
</ul>

// Leetcode Basic Template
#define ll long long
#define pb push_back
#define f(i,a,b) for(int i=a; i<=b; i++)
#define pi pair<int,int>
#define vi vector<int>
#define vpi vector<pair<int,int>>
#define F first
#define S second
#define L left
#define R right
#define all(v) v.begin(), v.end()
class Solution {
public:
    int subarrayBitwiseORs(vector<int>& arr) {
        int n = arr.size();
        set<int> s;
        vector<vi> v(n);
        v[0].pb(arr[0]);
        s.insert(arr[0]);
        f(i,1,n-1) {
            s.insert(arr[i]);
            v[i].pb(arr[i]);
            for(auto it:v[i-1]) {
                int val = (arr[i] | it);
                if(val>v[i].back()) {
                    s.insert(val);
                    v[i].pb(val);
                }
            }
        }

        return s.size();
    }
};
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