<p>You are given a 2D integer array <code>intervals</code>, where <code>intervals[i] = [left<sub>i</sub>, right<sub>i</sub>]</code> describes the <code>i<sup>th</sup></code> interval starting at <code>left<sub>i</sub></code> and ending at <code>right<sub>i</sub></code> <strong>(inclusive)</strong>. The <strong>size</strong> of an interval is defined as the number of integers it contains, or more formally <code>right<sub>i</sub> - left<sub>i</sub> + 1</code>.</p>

<p>You are also given an integer array <code>queries</code>. The answer to the <code>j<sup>th</sup></code> query is the <strong>size of the smallest interval</strong> <code>i</code> such that <code>left<sub>i</sub> &lt;= queries[j] &lt;= right<sub>i</sub></code>. If no such interval exists, the answer is <code>-1</code>.</p>

<p>Return <em>an array containing the answers to the queries</em>.</p>

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

<pre>
<strong>Input:</strong> intervals = [[1,4],[2,4],[3,6],[4,4]], queries = [2,3,4,5]
<strong>Output:</strong> [3,3,1,4]
<strong>Explanation:</strong> The queries are processed as follows:
- Query = 2: The interval [2,4] is the smallest interval containing 2. The answer is 4 - 2 + 1 = 3.
- Query = 3: The interval [2,4] is the smallest interval containing 3. The answer is 4 - 2 + 1 = 3.
- Query = 4: The interval [4,4] is the smallest interval containing 4. The answer is 4 - 4 + 1 = 1.
- Query = 5: The interval [3,6] is the smallest interval containing 5. The answer is 6 - 3 + 1 = 4.
</pre>

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

<pre>
<strong>Input:</strong> intervals = [[2,3],[2,5],[1,8],[20,25]], queries = [2,19,5,22]
<strong>Output:</strong> [2,-1,4,6]
<strong>Explanation:</strong> The queries are processed as follows:
- Query = 2: The interval [2,3] is the smallest interval containing 2. The answer is 3 - 2 + 1 = 2.
- Query = 19: None of the intervals contain 19. The answer is -1.
- Query = 5: The interval [2,5] is the smallest interval containing 5. The answer is 5 - 2 + 1 = 4.
- Query = 22: The interval [20,25] is the smallest interval containing 22. The answer is 25 - 20 + 1 = 6.
</pre>

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

<ul>
	<li><code>1 &lt;= intervals.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= queries.length &lt;= 10<sup>5</sup></code></li>
	<li><code>intervals[i].length == 2</code></li>
	<li><code>1 &lt;= left<sub>i</sub> &lt;= right<sub>i</sub> &lt;= 10<sup>7</sup></code></li>
	<li><code>1 &lt;= queries[j] &lt;= 10<sup>7</sup></code></li>
</ul>

class Solution:
    def minInterval(self, intervals: List[List[int]], queries: List[int]) -> List[int]:
        queries_copy=queries.copy()
        intervals.sort()
        queries.sort()
        #print(intervals)
        #print(queries)
        #print(30*'-')
        n,q= len(intervals), len(queries)
        idx=0
        heap=[]
        idx=0
        while idx<n:
            l,r=intervals[idx]
            if l<=queries[0] and queries[0]<=r:
                heappush(heap, (r-l+1, r))
                idx+=1
            else:
                break
        
        #print(heap)
        ans={x:-1 for x in queries}
        for j in range(q):
            x=queries[j]
            if j==0:
                if heap:
                    s,r = heappop(heap)
                    while r<x and heap:
                        s,r = heappop(heap)
                    if x<=r:
                        ans[x]=s
                        heappush(heap, (s,r))
                #print(ans[x])
                #print(20*'-')
            else:
                while idx<n:
                    l,r = intervals[idx]
                    if x>r:
                        idx+=1
                    elif l<=x and x<=r:
                        heappush(heap, (r-l+1, r))
                        idx+=1
                    else:
                        break
                if heap:
                    s,r = heappop(heap)
                    while r<x and heap:
                        s,r = heappop(heap)
                    if x<=r:
                        ans[x]=s
                        heappush(heap, (s,r))
                #print(ans[x])
                #print(20*'-')
            #print(heap)
        return [ans[x] for x in queries_copy]


                




        
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