Mr. Kitayuta's Colorful Graph
Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Description

Mr. Kitayuta has just bought an undirected graph consisting of n vertices and m edges. The vertices of the graph are numbered from  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 ( ≤ n ≤ ,  ≤ m ≤ ), denoting the number of the vertices and the number of the edges, respectively.

The next m lines contain space-separated three integers — ai, bi ( ≤ ai < bi ≤ n) and ci ( ≤ 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 ( ≤ q ≤ ), denoting the number of the queries.

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

Output

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

Sample Input

Input

Output

Input

Output

Hint

Let's consider the first sample.

 The figure above shows the first sample.

Vertex  and vertex  are connected by color  and .

Vertex  and vertex  are connected by color .

Vertex  and vertex  are not connected by any single color.

题目还是比较水，很简单的并查集，思路清晰就够了；
AC代码：

#include"iostream"

#include"cstdio"

#include"cstring"

#include"cmath"

#include"algorithm"

using namespace std;

const int MX=110;

int CL[101];

struct nde {

	int	p[MX];	//记录每个颜色下的根

} cmap[MX];

struct node {

	int a,b,c; //连接的点 a b 和颜色c

} side[MX];

bool cmps(node a,node b) {

	return a.c<b.c;

}

struct nod {

	int u,v;	//查询组

} Que[MX];

int find(int x,int c) {

	return cmap[c].p[x]==x?x:(cmap[c].p[x]=find(cmap[c].p[x],c)); //找到在颜色C下的根。

}

void init() {				//清空并初始化整个颜色图。

	for(int i=1; i<=100; i++)

		for(int j=1; j<=100; j++) {

			cmap[i].p[j]=j;

		}

}

int main() {

	int n,m,q;

	while(~scanf("%d%d",&n,&m)) {

		init();

		for(int i=1; i<=m; i++) {		//把整个输入先全部记录下

			scanf("%d%d%d",&side[i].a,&side[i].b,&side[i].c);

		}

		sort(side+1,side+1+m,cmps);//按照颜色排序；

		int tot=1;		//tot 记录颜色的种类

		CL[tot]=side[1].c;//记录下出现过的颜色

		for(int i=1; i<=m; i++) {

			if(side[i].c!=CL[tot]) {	//如果下一个颜色与上一个颜色不同记录进数组  CL

				CL[++tot]=side[i].c;

			}

			int rt1=find(side[i].a,side[i].c);  	//找到在颜色 C 下的根

			int rt2=find(side[i].b,side[i].c);

			if(rt1!=rt2) {

				cmap[side[i].c].p[rt2]=rt1;	//在颜色 C下将两点连接，更新根

			}

		}

		scanf("%d",&q);

		for(int i=1; i<=q; i++) {

			scanf("%d%d",&Que[i].u,&Que[i].v);	//单组输入查询

			int ans=0;

			for(int j=1; j<=tot; j++) {			//每个出现过的颜色下都查询一次

				int rt1=find(Que[i].u,CL[j]);

				int rt2=find(Que[i].v,CL[j]);

				if(rt1==rt2)ans++;				//如果是一个联通块答案加一

			}

			printf("%d\n",ans);					//输出该组的答案

		}

	}

	return 0;

}