<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]:
        ans = [-1]*len(queries)

        intervals.sort()
        qidxs = list(range(len(queries)))
        qidxs.sort(key=lambda i: queries[i])

        k = 0 # next interval index to push
        ivals = [] # (size, start, end) for all intervals containing next query, min heap by size
        for i in qidxs:
            q = queries[i]

            while ivals and ivals[0][2] < q: heappop(ivals)
            
            while k < len(intervals) and intervals[k][0] <= q:
                if intervals[k][1] >= q:
                    # skip intervals that start later, but have already ended
                    # (we'd pop them to maintain RI anyway)
                    heappush(ivals, (intervals[k][1]-intervals[k][0]+1, intervals[k][0], intervals[k][1]))
                k += 1

            if ivals:
                ans[i] = ivals[0][0]

        return ans
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