Mr. Kitayuta has just bought an undirected graph consisting of n vertices and m edges. The vertices of the graph are numbered from 1 to n. Each edge, namely edge i, has a color c_i, connecting vertex a_i and b_i.

Mr. Kitayuta wants you to process the following q queries.

In the i-th query, he gives you two integers — u_i and v_i.

Find the number of the colors that satisfy the following condition: the edges of that color connect vertex u_i and vertex v_i directly or indirectly.

The first line of the input contains space-separated two integers — n and m (2 ≤ n ≤ 100, 1 ≤ m ≤ 100), denoting the number of the vertices and the number of the edges, respectively.

The next m lines contain space-separated three integers — a_i, b_i (1 ≤ a_i < b_i ≤ n) and c_i (1 ≤ c_i ≤ m). Note that there can be multiple edges between two vertices. However, there are no multiple edges of the same color between two vertices, that is, if i ≠ j,(a_i, b_i, c_i) ≠ (a_j, b_j, c_j).

The next line contains a integer — q (1 ≤ q ≤ 100), denoting the number of the queries.

Then follows q lines, containing space-separated two integers — u_i and v_i (1 ≤ u_i, v_i ≤ n). It is guaranteed that u_i ≠ v_i.

For each query, print the answer in a separate line.

Let's consider the first sample.

 The figure above shows the first sample.

• Vertex 1 and vertex 2 are connected by color 1 and 2.
• Vertex 3 and vertex 4 are connected by color 3.
• Vertex 1 and vertex 4 are not connected by any single color.

#include<bits/stdc++.h>

#define maxx 105

using namespace std;

vector <int>edg[maxx][maxx];

bool vis[maxx];

int ans;

int x,y;

bool bfs(int c)

{

    queue<int> q;

    q.push(x);

    int now;

    memset(vis,,sizeof(vis));

    while(!q.empty())

    {

        now=q.front();

        if(now==y) return true;

        q.pop();

        for(int i=;i<edg[c][now].size();i++)

        {

            int then=edg[c][now][i];

            if(vis[then]) continue;

            vis[then]=;

            q.push(then);

        }

    }

    return false;

}

int main()

{

    int n,m;

    scanf("%d%d",&n,&m);

    int a,b,c;

    for(int i=;i<m;i++)

    {

        scanf("%d%d%d",&a,&b,&c);

        edg[c][b].push_back(a);

        edg[c][a].push_back(b);

    }

    int q;

    scanf("%d",&q);

    for(int i=;i<q;i++)

    {

        ans=;

        scanf("%d%d",&x,&y);

        for(int i=;i<m;i++)

            if(bfs(i+)) ans++;

         printf("%d\n",ans);

    }

}

这道题不难理解，用bfs，保存每种颜色的边的关联的边，然后对每种颜色的边bfs，如果能够找到符合起点终点的边，结果就加一，知道找完所有的颜色的边。

据说这题还可以用并查集做。有机会看一下。学渣现在正处于学习阶段。