<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 Solution:
    def minimumCost(self, n: int, edges: List[List[int]], query: List[List[int]]) -> List[int]:
        # initialize the dad array
        dad = [-1] * n
        
        #print(f'Dad = {dad}')
        # for the disjoint set structure
        for (u, v, _) in edges:
            #print(f'Union {u} to {v}')
            self.union(u, v, dad)
            #print(f'Dad = {dad}')
        
        # find total of all edges in each disjoint set
        component = dict()
        for (u, v, w) in edges:
            root = self.find(u, dad)
            if root not in component:
                component[root] = w
            else:
                component[root] = component[root] & w
        
        # you can never increase the cost by and'ing with another edge
        # having more than one instance of an edge in the path does not alter the cost
        # minimum cost is found by and'ing all of the edges 
    
        #print(f'Dad ={dad}')
        result = list()
        for (u, v) in query:
            root_u = self.find(u, dad)
            root_v = self.find(v, dad)
            
            if root_u == root_v:
                # they're in the same component, therefore we must "and" up all of the edges in that component
                result.append(component[root_u])
            else:
                result.append(-1)
        return result
    
       
    def union(self, u, v, dad):
        root_u = self.find(u, dad)
        root_v = self.find(v, dad)
        
        #print(f'root of {u} = {root_u}, root of {v} = {root_v}')
        if root_u == root_v:
            return
        
        val_u = dad[root_u]
        val_v = dad[root_v]
        
        # component containing u has more elements
        if val_u < val_v:
            temp = dad[root_v]
            dad[root_v] = root_u
            dad[root_u] = temp + dad[root_u]
        else:
            temp = dad[root_u]
            dad[root_u] = root_v
            dad[root_v] = temp + dad[root_v]
            
    def find(self, v, dad):
        cur = v
        #print(f'Dad = {dad}')
        # find the root
        while dad[cur] >= 0:
            cur = dad[cur]
        
        root = cur
        cur = v
        # path compression
        while dad[cur] >= 0:
            temp = dad[cur]
            dad[cur] = root
            cur = temp
        
        return root
96700