<p>There is an undirected weighted graph with <code>n</code> vertices labeled from <code>0</code> to <code>n - 1</code>.</p>

<p>You are given the integer <code>n</code> and an array <code>edges</code>, where <code>edges[i] = [u<sub>i</sub>, v<sub>i</sub>, w<sub>i</sub>]</code> indicates that there is an edge between vertices <code>u<sub>i</sub></code> and <code>v<sub>i</sub></code> with a weight of <code>w<sub>i</sub></code>.</p>

<p>A walk on a graph is a sequence of vertices and edges. The walk starts and ends with a vertex, and each edge connects the vertex that comes before it and the vertex that comes after it. It&#39;s important to note that a walk may visit the same edge or vertex more than once.</p>

<p>The <strong>cost</strong> of a walk starting at node <code>u</code> and ending at node <code>v</code> is defined as the bitwise <code>AND</code> of the weights of the edges traversed during the walk. In other words, if the sequence of edge weights encountered during the walk is <code>w<sub>0</sub>, w<sub>1</sub>, w<sub>2</sub>, ..., w<sub>k</sub></code>, then the cost is calculated as <code>w<sub>0</sub> &amp; w<sub>1</sub> &amp; w<sub>2</sub> &amp; ... &amp; w<sub>k</sub></code>, where <code>&amp;</code> denotes the bitwise <code>AND</code> operator.</p>

<p>You are also given a 2D array <code>query</code>, where <code>query[i] = [s<sub>i</sub>, t<sub>i</sub>]</code>. For each query, you need to find the minimum cost of the walk starting at vertex <code>s<sub>i</sub></code> and ending at vertex <code>t<sub>i</sub></code>. If there exists no such walk, the answer is <code>-1</code>.</p>

<p>Return <em>the array </em><code>answer</code><em>, where </em><code>answer[i]</code><em> denotes the <strong>minimum</strong> cost of a walk for query </em><code>i</code>.</p>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">n = 5, edges = [[0,1,7],[1,3,7],[1,2,1]], query = [[0,3],[3,4]]</span></p>

<p><strong>Output:</strong> <span class="example-io">[1,-1]</span></p>

<p><strong>Explanation:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2024/01/31/q4_example1-1.png" style="padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem; width: 351px; height: 141px;" />
<p>To achieve the cost of 1 in the first query, we need to move on the following edges: <code>0-&gt;1</code> (weight 7), <code>1-&gt;2</code> (weight 1), <code>2-&gt;1</code> (weight 1), <code>1-&gt;3</code> (weight 7).</p>

<p>In the second query, there is no walk between nodes 3 and 4, so the answer is -1.</p>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">n = 3, edges = [[0,2,7],[0,1,15],[1,2,6],[1,2,1]], query = [[1,2]]</span></p>

<p><strong>Output:</strong> <span class="example-io">[0]</span></p>

<p><strong>Explanation:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2024/01/31/q4_example2e.png" style="padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem; width: 211px; height: 181px;" />
<p>To achieve the cost of 0 in the first query, we need to move on the following edges: <code>1-&gt;2</code> (weight 1), <code>2-&gt;1</code> (weight 6), <code>1-&gt;2</code> (weight 1).</p>
</div>

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

<ul>
	<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= edges.length &lt;= 10<sup>5</sup></code></li>
	<li><code>edges[i].length == 3</code></li>
	<li><code>0 &lt;= u<sub>i</sub>, v<sub>i</sub> &lt;= n - 1</code></li>
	<li><code>u<sub>i</sub> != v<sub>i</sub></code></li>
	<li><code>0 &lt;= w<sub>i</sub> &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= query.length &lt;= 10<sup>5</sup></code></li>
	<li><code>query[i].length == 2</code></li>
	<li><code>0 &lt;= s<sub>i</sub>, t<sub>i</sub> &lt;= n - 1</code></li>
	<li><code>s<sub>i</sub> !=&nbsp;t<sub>i</sub></code></li>
</ul>

class DSU:
    parent = {}
    size = {}
    def make_set(self, node: int):
        self.parent[node] = node
        self.size[node] = 1
    def find_set(self, node: int):
        if self.parent[node] == node:
            return node
        self.parent[node] = self.find_set(self.parent[node])
        return self.parent[node]
    def union_sets(self, left: int, right: int):
        set1 = self.find_set(left)
        set2 = self.find_set(right)
        if set1 != set2:
            if self.size[set1] < self.size[set2]:
                set1, set2 = set2, set1
            self.parent[set2] = set1
            self.size[set1] += self.size[set2]
            self.size[set2] = 0


class Solution:
    def minimumCost(self, n: int, edges: List[List[int]], query: List[List[int]]) -> List[int]:
        dsu = DSU()
        for i in range(n):
            dsu.make_set(i)
        for left, right, _ in edges:
            dsu.union_sets(left, right)
        
        set_and = {}
        for [left, _, weight] in edges:
            s = dsu.find_set(left)
            if s not in set_and:
                set_and[s] = weight
            set_and[s] &= weight

        res = []
        for [start, end] in query:
            s1 = dsu.find_set(start)
            s2 = dsu.find_set(end)
            if s1 != s2:
                res.append(-1)
            else:
                res.append(set_and[s1])

        return res
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