【Codeforces-505B】Mr. Kitayuta's Colorful Graph（二维并查集）

Mr. Kitayuta's Colorful Graph

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

Copy

4 5 1 2 1 1 2 2 2 3 1 2 3 3 2 4 3 3 1 2 3 4 1 4

output

Copy

2 1 0

input

Copy

5 7 1 5 1 2 5 1 3 5 1 4 5 1 1 2 2 2 3 2 3 4 2 5 1 5 5 1 2 5 1 5 1 4

output

Copy

1 1 1 1 2

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.

题意：

给你n个点m条边，然后告诉你m条边，链接a,b并且颜色是c。然后又q个询问，问同一种颜色的路径连接顶点 ui 和 顶点 vi 的数量。

思路：

颜色数和点的个数不大，因此可以对每一种颜色维护一个并查集，然后查询一对点是，遍历所有颜色的并查集。

#include<bits/stdc++.h> using namespace std; typedef long long ll; int f[110][110]; void init(){ for(int i=0;i<=105;i++) { for(int j=0;j<=105;j++) { f[i][j]=j; } } } int getf(int i,int x) { if(f[i][x]!=x) f[i][x]=getf(i,f[i][x]); return f[i][x]; } void merge(int u,int v,int c) { int t1=getf(c,u); int t2=getf(c,v); if(t1!=t2){ f[c][t1]=t2; } } int main() { init(); int n,m; scanf("%d%d",&n,&m); for(int i=0;i<m;i++) { int a,b,c; scanf("%d%d%d",&a,&b,&c); merge(a,b,c); } int q; scanf("%d",&q); while(q--) { int u,v; scanf("%d%d",&u,&v); int ans=0; for(int i=1;i<=m;i++) { if(getf(i,u)==getf(i,v)) ans++; } printf("%d\n",ans); } return 0; }