There is an undirected weighted graph with n vertices labeled from 0 to n - 1.
You are given the integer n and an array edges, where edges[i] = [ui, vi, wi] indicates that there is an edge between vertices ui and vi with a weight of wi.
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's important to note that a walk may visit the same edge or vertex more than once.
\n\nThe cost of a walk starting at node u and ending at node v is defined as the bitwise AND of the weights of the edges traversed during the walk. In other words, if the sequence of edge weights encountered during the walk is w0, w1, w2, ..., wk, then the cost is calculated as w0 & w1 & w2 & ... & wk, where & denotes the bitwise AND operator.
You are also given a 2D array query, where query[i] = [si, ti]. For each query, you need to find the minimum cost of the walk starting at vertex si and ending at vertex ti. If there exists no such walk, the answer is -1.
Return the array answer, where answer[i] denotes the minimum cost of a walk for query i.
\n
Example 1:
\n\nInput: n = 5, edges = [[0,1,7],[1,3,7],[1,2,1]], query = [[0,3],[3,4]]
\n\nOutput: [1,-1]
\n\nExplanation:
\n![\"\"](\"https://assets.leetcode.com/uploads/2024/01/31/q4_example1-1.png\")
\nTo achieve the cost of 1 in the first query, we need to move on the following edges: 0->1 (weight 7), 1->2 (weight 1), 2->1 (weight 1), 1->3 (weight 7).
In the second query, there is no walk between nodes 3 and 4, so the answer is -1.
\n\nExample 2:
\nInput: n = 3, edges = [[0,2,7],[0,1,15],[1,2,6],[1,2,1]], query = [[1,2]]
\n\nOutput: [0]
\n\nExplanation:
\n![\"\"](\"https://assets.leetcode.com/uploads/2024/01/31/q4_example2e.png\")
\nTo achieve the cost of 0 in the first query, we need to move on the following edges: 1->2 (weight 1), 2->1 (weight 6), 1->2 (weight 1).
\n
Constraints:
\n\n1 <= n <= 1050 <= edges.length <= 105edges[i].length == 30 <= ui, vi <= n - 1ui != vi0 <= wi <= 1051 <= query.length <= 105query[i].length == 20 <= si, ti <= n - 1给你一个 n 个节点的带权无向图，节点编号为 0 到 n - 1 。
给你一个整数 n 和一个数组 edges ，其中 edges[i] = [ui, vi, wi] 表示节点 ui 和 vi 之间有一条权值为 wi 的无向边。
在图中，一趟旅途包含一系列节点和边。旅途开始和结束点都是图中的节点，且图中存在连接旅途中相邻节点的边。注意，一趟旅途可能访问同一条边或者同一个节点多次。
\n\n如果旅途开始于节点 u ，结束于节点 v ，我们定义这一趟旅途的 代价 是经过的边权按位与 AND 的结果。换句话说，如果经过的边对应的边权为 w0, w1, w2, ..., wk ，那么代价为w0 & w1 & w2 & ... & wk ，其中 & 表示按位与 AND 操作。
给你一个二维数组 query ，其中 query[i] = [si, ti] 。对于每一个查询，你需要找出从节点开始 si ，在节点 ti 处结束的旅途的最小代价。如果不存在这样的旅途，答案为 -1 。
返回数组 answer ，其中 answer[i] 表示对于查询 i 的 最小 旅途代价。
\n\n
示例 1：
\n\n输入：n = 5, edges = [[0,1,7],[1,3,7],[1,2,1]], query = [[0,3],[3,4]]
\n\n输出：[1,-1]
\n\n解释：
\n\n
第一个查询想要得到代价为 1 的旅途，我们依次访问：0->1（边权为 7 ）1->2 （边权为 1 ）2->1（边权为 1 ）1->3 （边权为 7 ）。
第二个查询中，无法从节点 3 到节点 4 ，所以答案为 -1 。
\n\n示例 2：
\n输入：n = 3, edges = [[0,2,7],[0,1,15],[1,2,6],[1,2,1]], query = [[1,2]]
\n\n输出：[0]
\n\n解释：
\n\n
第一个查询想要得到代价为 0 的旅途，我们依次访问：1->2（边权为 1 ），2->1（边权 为 6 ），1->2（边权为 1 ）。
\n\n
提示：
\n\n1 <= n <= 1050 <= edges.length <= 105edges[i].length == 30 <= ui, vi <= n - 1ui != vi0 <= wi <= 1051 <= query.length <= 105query[i].length == 20 <= si, ti <= n - 1u and v are not connected then the answer is -1, otherwise we can use all the edges from the connected component where both belong to."
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