Mr. Kitayuta's Colorful Graph

数据量很小用搜索或并查集应该都能过

题目要求：

从一个点到另一个点可以经过多少条不同的边

思路：

为每一条不同编号的边建一个并查集，查看有多少条边可以使这两个点有共同的父节点

                                        B. Mr. Kitayuta's Colorful Graph

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 vidirectly 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

4 51 2 11 2 22 3 12 3 32 4 331 23 41 4

output

210

input

5 71 5 12 5 13 5 14 5 11 2 22 3 23 4 251 55 12 51 51 4

output

11112

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.

#include<stdio.h>#include<string.h>#define maxn 105int pre[maxn][maxn];void Union(int x,int y,int z);int find(int a,int b);int main() {int n,m,q;memset(pre,0,sizeof(pre));scanf("%d%d",&n,&m);for(int i=1;i<=m;i++){for(int j=1;j<=n;j++){pre[i][j]=j;}}for(int i=0;i<m;i++){int a,b,c;scanf("%d%d%d",&a,&b,&c);Union(a,b,c);}scanf("%d",&q);for(int i=0;i<q;i++){int a,b,ans=0;scanf("%d%d",&a,&b);for(int j=1;j<=m;j++){int fa,fb;fa=find(a,j);fb=find(b,j);if(fa==fb)ans++;}printf("%d\n",ans);}return 0;}void Union(int x,int y,int z){int fx,fy;fx=find(x,z);fy=find(y,z);if(fx!=fy)pre[z][fx]=fy;}int find(int a,int b){int r=a;while(r!=pre[b][r])r=pre[b][r];while(a!=pre[b][a]){int s=a;a=pre[b][a];pre[b][s]=r;}return r;}