<p>A <strong>good meal</strong> is a meal that contains <strong>exactly two different food items</strong> with a sum of deliciousness equal to a power of two.</p>

<p>You can pick <strong>any</strong> two different foods to make a good meal.</p>

<p>Given an array of integers <code>deliciousness</code> where <code>deliciousness[i]</code> is the deliciousness of the <code>i<sup>​​​​​​th</sup>​​​​</code>​​​​ item of food, return <em>the number of different <strong>good meals</strong> you can make from this list modulo</em> <code>10<sup>9</sup> + 7</code>.</p>

<p>Note that items with different indices are considered different even if they have the same deliciousness value.</p>

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

<pre>
<strong>Input:</strong> deliciousness = [1,3,5,7,9]
<strong>Output:</strong> 4
<strong>Explanation: </strong>The good meals are (1,3), (1,7), (3,5) and, (7,9).
Their respective sums are 4, 8, 8, and 16, all of which are powers of 2.
</pre>

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

<pre>
<strong>Input:</strong> deliciousness = [1,1,1,3,3,3,7]
<strong>Output:</strong> 15
<strong>Explanation: </strong>The good meals are (1,1) with 3 ways, (1,3) with 9 ways, and (1,7) with 3 ways.</pre>

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

<ul>
	<li><code>1 &lt;= deliciousness.length &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= deliciousness[i] &lt;= 2<sup>20</sup></code></li>
</ul>

class Solution {
public:
    int countPairs(vector<int>& deliciousness) {
        ios::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);
        
        long ans=0;
        int n=deliciousness.size();
        sort(deliciousness.begin(), deliciousness.end());
        
        vector<pair<int,long>> freq;
        int d=deliciousness[0], f=1;
        for (int i=1; i<n; i++) {
            if (deliciousness[i-1]!=deliciousness[i]) {
                freq.push_back({deliciousness[i-1], f});
                f=1;
            }
            else {
                f+=1;
            }
        }
        freq.push_back({deliciousness[n-1], f});
        
        for (int i=0, m=freq.size(); i<m; i++) {
            for (int j=i; j<m; j++) {
                int d=freq[i].first+freq[j].first;
                if (d && (d&(d-1))==0) {
                    if (i==j) {
                        ans=(ans+freq[i].second*(freq[i].second-1)/2)%1000000007;
                    }
                    else {
                        ans=(ans+freq[i].second*freq[j].second)%1000000007;
                    }
                }   
            }
        }
        return ans;
    }
};
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