<p>You are given an undirected graph defined by an integer <code>n</code>, the number of nodes, and a 2D integer array <code>edges</code>, the edges in the graph, where <code>edges[i] = [u<sub>i</sub>, v<sub>i</sub>]</code> indicates that there is an <strong>undirected</strong> edge between <code>u<sub>i</sub></code> and <code>v<sub>i</sub></code>. You are also given an integer array <code>queries</code>.</p>

<p>Let <code>incident(a, b)</code> be defined as the <strong>number of edges</strong> that are connected to <strong>either</strong> node <code>a</code> or <code>b</code>.</p>

<p>The answer to the <code>j<sup>th</sup></code> query is the <strong>number of pairs</strong> of nodes <code>(a, b)</code> that satisfy <strong>both</strong> of the following conditions:</p>

<ul>
	<li><code>a &lt; b</code></li>
	<li><code>incident(a, b) &gt; queries[j]</code></li>
</ul>

<p>Return <em>an array </em><code>answers</code><em> such that </em><code>answers.length == queries.length</code><em> and </em><code>answers[j]</code><em> is the answer of the </em><code>j<sup>th</sup></code><em> query</em>.</p>

<p>Note that there can be <strong>multiple edges</strong> between the same two nodes.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/06/08/winword_2021-06-08_00-58-39.png" style="width: 529px; height: 305px;" />
<pre>
<strong>Input:</strong> n = 4, edges = [[1,2],[2,4],[1,3],[2,3],[2,1]], queries = [2,3]
<strong>Output:</strong> [6,5]
<strong>Explanation:</strong> The calculations for incident(a, b) are shown in the table above.
The answers for each of the queries are as follows:
- answers[0] = 6. All the pairs have an incident(a, b) value greater than 2.
- answers[1] = 5. All the pairs except (3, 4) have an incident(a, b) value greater than 3.
</pre>

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

<pre>
<strong>Input:</strong> n = 5, edges = [[1,5],[1,5],[3,4],[2,5],[1,3],[5,1],[2,3],[2,5]], queries = [1,2,3,4,5]
<strong>Output:</strong> [10,10,9,8,6]
</pre>

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

<ul>
	<li><code>2 &lt;= n &lt;= 2 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= edges.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= u<sub>i</sub>, v<sub>i</sub> &lt;= n</code></li>
	<li><code>u<sub>i </sub>!= v<sub>i</sub></code></li>
	<li><code>1 &lt;= queries.length &lt;= 20</code></li>
	<li><code>0 &lt;= queries[j] &lt; edges.length</code></li>
</ul>


class Solution:
    def countPairs(self, n: int, edges: List[List[int]], queries: List[int]) -> List[int]:
        cnter={}
        self_connected=Counter()
        for i in range(1,n+1):cnter[i]=0
        for a,b in edges:
            if a>b:a,b=b,a
            cnter[a]+=1
            cnter[b]+=1
            self_connected[a,b]+=1

        sl=sorted(cnter.values())
        res=[]
        for q in queries:
            ans=0
            for itr,i in enumerate(sl):
                pos=bisect_right(sl,q-i)
                if pos>itr:ans+=len(sl)-pos
                else:
                    rem=len(sl)-itr
                    ans+=comb(rem,2)
                    break

            for (a,b),v in self_connected.items():
                if cnter[a]+cnter[b] > q and cnter[a]+cnter[b] - v <=q: 
                    ans-=1
            
            res.append(ans)
        
        return res
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