Mr. Kitayuta's Colorful Graph

原题链接 http://acm.hdu.edu.cn/showproblem.php?pid=3999

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 ci, connecting vertex ai and bi.

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

In the i-th query, he gives you two integers — ui and vi.

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

Input

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 — ai, bi (1 ≤ ai < bi ≤ n) and ci (1 ≤ ci ≤ 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, (ai, bi, ci) ≠ (aj, bj, cj).

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 — ui and vi (1 ≤ ui, vi ≤ n). It is guaranteed that ui ≠ vi.

Output

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

Examples

Input

Output

2
1
0

Input

Output

Note

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.
有m个颜色，在有向图中构建对点进行找祖归宗，输出联通线路的个数；
开始想到的便是并查集，不过不是太熟练，样例卡了几次，做了出来，就是并查集模板，但是要变成二维数组，最后进行查询即可。

#include<stdio.h>
#include<string.h>

int mapp[200][200];

void init()//初始化数组，为构建地图做准备
{
    for(int i=1; i<=100; i++)
    {
        for(int j=1; j<=100; j++)
        {
            mapp[i][j]=i;
        }
    }
}

int getf(int u,int v)//路径压缩
{
    if(mapp[u][v]==u)
    {
        return u;
    }
    else
    {
        return mapp[u][v]=getf(mapp[u][v],v);
    }
}

void merge(int u,int v,int w)//合并两子集合,找祖归宗
{
    int t1=getf(u,w);
    int t2=getf(v,w);
    if(t1!=t2)
    {
        mapp[t1][w]=t2;
    }
}

int main()
{
    int n,m;
    while(~scanf("%d %d",&n,&m))
    {
        init();
        int a,b,c;
        for(int i=1; i<=m; i++)
        {
            scanf("%d%d%d",&a,&b,&c);
            merge(a,b,c);
        }
        int q,u,v;
        scanf("%d",&q);//q行开始输入ui与vi
        for(int i=0; i<q; i++)
        {
            scanf("%d %d",&u,&v);
            int ans=0;
            for(int i=1; i<=m; i++)
            {
                if(getf(u,i)==getf(v,i))//如果有祖宗相同的，便是一条联通路线
                {
                    ans++;
                }
            }
            printf("%d\n",ans);
        }
    }
    return 0;
}

Mr. Kitayuta's Colorful Graph

Mr. Kitayuta's Colorful Graph

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