Mr. Kitayuta has just bought an undirected graph with 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 ≤ 105, 1 ≤ m ≤ 105), 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 ≤ 105), 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 5
1 2 1
1 2 2
2 3 1
2 3 3
2 4 3
3
1 2
3 4
1 4


Output

2
1
0


Input

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

1
1
1
1
2

Note

Let's consider the first sample. 

<image> 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>
using namespace std;
int dx[] = {-1, -1, -1, 0, 0, 1, 1, 1};
int dy[] = {-1, 0, 1, -1, 1, -1, 0, 1};
bool isPath = false;
void bfsConfirm(vector<pair<int, int> > graph[], int stNode, int enNode,
                bool visited[], int color) {
  visited[stNode] = true;
  if (enNode == stNode) {
    isPath = true;
    return;
  }
  for (auto neighbours : graph[stNode]) {
    int c = neighbours.second;
    int node = neighbours.first;
    if (c == color && visited[node] == false) {
      bfsConfirm(graph, node, enNode, visited, color);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t != 0) {
    t--;
    int n, m;
    cin >> n >> m;
    vector<pair<int, int> > graph[n + 1];
    while (m != 0) {
      m--;
      int a, b, c;
      cin >> a >> b >> c;
      graph[a].push_back(make_pair(b, c));
      graph[b].push_back(make_pair(a, c));
    }
    int queries;
    cin >> queries;
    while (queries != 0) {
      queries--;
      int st, en;
      cin >> st >> en;
      set<int> answers;
      bool visited[n + 1];
      for (auto neighbours : graph[st]) {
        isPath = false;
        memset(visited, false, sizeof(visited));
        int col = neighbours.second;
        int stNode = neighbours.first;
        bfsConfirm(graph, stNode, en, visited, col);
        if (isPath == true) {
          answers.insert(col);
        }
      }
      cout << answers.size() << "\n";
    }
  }
  return 0;
}
